// Arup Guha
// 10/26/2013
// Solution to 2013 NCPC Problem B: Boiling Vegetables

import java.util.*;

public class vegetables {

	public static int[] values;
	public static int n;
	public static double ratio;

	public static void main(String[] args) {

		// Read in data.
		Scanner stdin = new Scanner(System.in);
		ratio = stdin.nextDouble();
		n = stdin.nextInt();

		values = new int[n];
		for (int i=0; i<n; i++)
			values[i] = stdin.nextInt();

		// Easy to sort from low to high.
		Arrays.sort(values);

		// Return first solution we find, since we are searching from low to high.
		System.out.println(solve(0, values[0], values[0], 0));
	}

	// Currently cutting the vegetable values[k] and our current min, max and cuts are passed in.
	public static int solve(int k, double curMin, double curMax, int curCuts) {

		// Here is our answer.
		if (k == n)
			return curCuts;

		// Possible number of cuts for this vegetable.
		double absMax = curMin/ratio;
		double absMin = curMax*ratio;
		int low = (int)Math.ceil(values[k]/absMax);
		int high = (int)(values[k]/absMin);

		// Special case when we start search.
		if (k == 0) {
			low = 1;
			high = values[0];
		}

		// Try i-1 cuts.
		for (int i=low; i<=high; i++) {
			double newL = values[k]/i;

			// Set these for the first iteration.
			if (k == 0) curMin = newL;
			if (k == 0) curMax = newL;

			// See if this arrangement works, if it does, return it.
			int ans = solve(k+1, Math.min(curMin, newL), Math.max(curMax, newL), curCuts+i-1);
			if (ans != -1)
				return ans;
		}

		// Nothing worked.
		return -1;
	}
}