// Arup Guha
// 9/5/2015
// Solution to 2015 UCF Locals Problem: Passing

// Note: I didn't come up with this, Travis explained the solution idea to me.
// In a nutshell, you create a natural graph induced by the input data. Any cut
// of that graph represents a valid split of classes. The minimum cut of that
// graph, as a consequence, represents the maximal expectation of students
// passing the course. So, we initially place all students in both classes,
// and then sever the graph into two partitions, one for each teacher.
// Of course, the minimum cut of a graph is always equal to the maximum flow.
// Finally, you can take that minimum cut value and subtract it from our
// ficticious situation where each student is in both classes. One final note -
// flow works best with ints, so it's much better to run it with ints.

import java.util.*;

public class passing {

	final public static double EPSILON = 1e-4;

	public static void main(String[] args) {

		Scanner stdin = new Scanner(System.in);
		int numCases = stdin.nextInt();

		// Go through each case.
		for (int loop=0; loop<numCases; loop++) {

			// Will store flow graph here, source(Matt) = n, sink(Sean) = n+1
			int n = stdin.nextInt();
			int[][] graph = new int[n+2][n+2];
			int sum = 0;

			// Restricting probabilities are chances each student passes Matt's class on their own.
			for (int i=0; i<n; i++) {
				graph[n][i] = convert(stdin.nextDouble());
				sum += graph[n][i];
			}

			// Do the same for Sean's classes on the other side.
			for (int i=0; i<n; i++) {
				graph[i][n+1] = convert(stdin.nextDouble());
				sum += graph[i][n+1];
			}

			// Now, read in all the "helpful" probabilities.
			int[][] data = new int[n][n];
			for (int i=0; i<n; i++) {
				for (int j=0; j<n; j++) {
					data[i][j] = convert(stdin.nextDouble());
					sum += data[i][j];
				}
			}

			// We must store the sum of person A helping person B and vice versa.
			for (int i=0; i<n; i++)
				for (int j=0; j<n; j++)
					graph[i][j] = data[i][j] + data[j][i];

			// Set up the network flow graph and output solution.
			networkflow myFlow = new networkflow(graph, n, n+1);
			System.out.printf("%.2f\n", (sum - myFlow.getMaxFlow() + EPSILON)/100);
		}
	}

	// Pre-condition: x is expressed to exactly 2 decimal places.
	// Returns 100x as an integer.
	public static int convert(double x) {
		return (int)(100*x+EPSILON);
	}
}

/*** My original network flow code follows. - it's much worse than Matt's!!! ***/

class Edge {

	private int capacity;
	private int flow;

	public Edge(int cap) {
		capacity = cap;
		flow = 0;
	}

	public int maxPushForward() {
		return capacity - flow;
	}

	public int maxPushBackward() {
		return flow;
	}

	public boolean pushForward(int moreflow) {

		// We can't push through this much flow.
		if (flow+moreflow > capacity)
			return false;

		// Push through.
		flow += moreflow;
		return true;
	}

	public boolean pushBack(int lessflow) {

		// Not enough to push back on.
		if (flow < lessflow)
			return false;

		flow -= lessflow;
		return true;
	}
}

class direction {

	public int prev;
	public boolean forward;

	public direction(int node, boolean dir) {
		prev = node;
		forward = dir;
	}

	public String toString() {
		if (forward)
			return "" + prev + "->";
		else
			return "" + prev + "<-";
	}
}

class networkflow {

	private Edge[][] adjMat;
	private int source;
	private int dest;

	// All positive entries in flows should represent valid flows
	// between vertices. All other entries must be 0 or negative.
	public networkflow(int[][] flows, int start, int end) {

		source = start;
		dest = end;
		adjMat = new Edge[flows.length][flows.length];

		for (int i=0; i<flows.length; i++) {
			for (int j=0; j<flows[i].length; j++) {

				// Fill in this flow.
				if (flows[i][j] > 0)
					adjMat[i][j] = new Edge(flows[i][j]);
				else
					adjMat[i][j] = null;
			}
		}
	}

	public ArrayList<direction> findAugmentingPath() {

		// This will store the previous node visited in the BFS.
		direction[] prev = new direction[adjMat.length];
		boolean[] inQueue = new boolean[adjMat.length];
		for (int i=0; i<inQueue.length; i++)
			inQueue[i] = false;

		// The source has no previous node.
		prev[source] = new direction(-1, true);

		LinkedList<Integer> bfs_queue = new LinkedList<Integer>();
		bfs_queue.offer(new Integer(source));
		inQueue[source] = true;

		// Our BFS will go until we clear out the queue.
		while (bfs_queue.size() > 0) {

			// Add all the new neighbors of the current node.
			Integer next = bfs_queue.poll();

			// Find all neighbors and add into the queue. These are forward edges.
			for (int i=0; i<adjMat.length; i++)
				if (!inQueue[i] && adjMat[next][i] != null && adjMat[next][i].maxPushForward() > 0) {
					bfs_queue.offer(new Integer(i));
					inQueue[i] = true;
					prev[i] = new direction(next, true);
				}

			// Now look for back edges.
			for (int i=0; i<adjMat.length; i++)
				if (!inQueue[i] && adjMat[i][next] != null && adjMat[i][next].maxPushBackward() > 0) {
					bfs_queue.offer(new Integer(i));
					inQueue[i] = true;
					prev[i] = new direction(next, false);
				}
		}

		// No augmenting path found.
		if (!inQueue[dest])
			return null;

		ArrayList<direction> path = new ArrayList<direction>();

		direction place = prev[dest];

		direction dummy = new direction(dest, true);
		path.add(dummy);

		// Build the path backwards.
		while (place.prev != -1) {
			path.add(place);
			place = prev[place.prev];
		}

		// Reverse it now.
		Collections.reverse(path);

		return path;
	}

	// Run the Max Flow Algorithm here.
	public int getMaxFlow() {

		int flow = 0;

		ArrayList<direction> nextpath = findAugmentingPath();

		// Loop until there are no more augmenting paths.
		while (nextpath != null) {

			// Check what the best flow through this path is.
			int this_flow = Integer.MAX_VALUE;
			for (int i=0; i<nextpath.size()-1; i++) {

				if (nextpath.get(i).forward) {
					this_flow = Math.min(this_flow, adjMat[nextpath.get(i).prev][nextpath.get(i+1).prev].maxPushForward());
				}
				else {
					this_flow = Math.min(this_flow, adjMat[nextpath.get(i+1).prev][nextpath.get(i).prev].maxPushBackward());
				}
			}

			// Now, put this flow through.
			for (int i=0; i<nextpath.size()-1; i++) {

				if (nextpath.get(i).forward) {
					adjMat[nextpath.get(i).prev][nextpath.get(i+1).prev].pushForward(this_flow);
				}
				else {
					adjMat[nextpath.get(i+1).prev][nextpath.get(i).prev].pushBack(this_flow);
				}
			}

			// Add this flow in and then get the next path.
			flow += this_flow;
			nextpath = findAugmentingPath();
		}

		return flow;
	}
}