// Arup Guha
// 1/16/2017
// Solution to 2014 NAQ Problem D: Flip Five

import java.util.*;

public class flipfive {

	// Directions for toggles (including self).
	final public static int[] DX = {0,-1,0,0,1};
	final public static int[] DY = {0,0,-1,1,0};

	// Dimensions for problem.
	final public static int R = 3;
	final public static int C = 3;

	public static void main(String[] args) {

		// Pre-process each solution - based on input directions, this is a one-to-one mapping,
		// no need to run BFS.
		int[] sol = new int[1<<(R*C)];
		for (int i=0; i<(1<<(R*C)); i++)
			sol[state(i)] = Integer.bitCount(i);

		// Process each case.
		Scanner stdin = new Scanner(System.in);
		int numCases = stdin.nextInt();
		for (int loop=0; loop<numCases; loop++) {

			// Read in grid.
			String s = stdin.next();
			for (int i=0; i<R-1; i++)
				s = s + stdin.next();

			// Output solution after conversion to bit mask (note - I flip this on vert and horiz axis, but that's okay.)
			System.out.println(sol[convert(s)]);
		}
	}

	// Returns binary value of s.
	public static int convert(String s) {

		int val = 0;

		// Go through each bit.
		for (int i=0; i<s.length(); i++) {

			// This always happens.
			val = (val << 1);

			// Just for bits that are on.
			if (s.charAt(i) == '*')
				val += 1;
		}

		return val;
	}

	public static int state(int bits) {

		int cur = 0;

		// Go through each bit that's on and flip each corresponding spot.
		for (int i=0; (1<<i)<=bits; i++)
			if ((bits & (1<<i)) != 0)
				cur = cur ^ flip(i);
		return cur;
	}

	public static int flip(int bit) {

		// Find bit.
		int myr = bit/C;
		int myc = bit%C;

		int res = 0;

		// Go through each spot, and add in mask for each one that's inbounds.
		for (int i=0; i<DX.length; i++)
			if (inbounds(myr+DX[i], myc+DY[i]))
				res += (1 << (C*(myr+DX[i]) + myc+DY[i]));

		return res;
	}

	public static boolean inbounds(int myr, int myc) {
		return myr >= 0 && myr < R && myc >= 0 && myc < C;
	}
}