Problem: You have a team of N people. For a particular task, you can pick any non-empty subset of people. The cost of having x people for the task is xk. 

Output the sum of costs over all non-empty subsets of people.

Input

Only line of input contains two integers N (1 ≤ N ≤ 109) representing total number of people and k (1 ≤ k ≤ 5000).

Output

Output the sum of costs for all non empty subsets modulo 109 + 7.

Examples

Input

1 1


Output

1


Input

3 2


Output

24

Note

In the first example, there is only one non-empty subset {1} with cost 11 = 1.

In the second example, there are seven non-empty subsets.

- {1} with cost 12 = 1

- {2} with cost 12 = 1

- {1, 2} with cost 22 = 4

- {3} with cost 12 = 1

- {1, 3} with cost 22 = 4

- {2, 3} with cost 22 = 4

- {1, 2, 3} with cost 32 = 9

The total cost is 1 + 1 + 4 + 1 + 4 + 4 + 9 = 24.

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.Scanner;
/*

1000000000 5000
642932262

 */
public class e {
	static long m = (long) (1e9 + 7);
	static long[][] choose;
	static long[] ps;
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt(), k = in.nextInt();
//		System.out.println("Choose");
		choose = new long[k + 1][];
		for (int x = 0; x < k + 1; x++) {
			choose[x] = new long[x + 1];
			choose[x][0] = choose[x][x] = 1;
			for (int y = 1; y < x; y++)
				choose[x][y] = (choose[x - 1][y - 1] + choose[x - 1][y]) % m;
		}
//		System.out.println("Ps");
		ps = new long[k+1];
		ps[0] = 1;
		for(int i=1;i<=k;i++)
			ps[i] = pow(i,k);
//		System.out.println("Solve");
		long[] ys = new long[k + 1];
		for (int nn = 0; nn <= k; nn++) {
			ys[nn] = (solve(nn) * mi(pow(2, nn - k))) % m;
		}
//		System.out.println("int");
		System.out.println((interpolate(ys, n) * pow(2, n - k)) % m);
	}
	
	static long solve(int n) {
		long y = 0;
		for (int i = 1; i <= n; i++) {
			y += choose[n][i] * ps[i];
			y %= m;
		}
		return y;
	}

	static long interpolate(long[] ys, int xi) {
		int n = ys.length;
		long result = 0;
		long[] mi = new long[n * 2 + 1];
		for (int d = -n; d <= n; d++) {
			mi[d + n] = mi(d+m);
		}
		long[] fact = new long[n+1];
		long[] ifact = new long[n+1];
		fact[0] = ifact[0] = 1;
		for(int i=1;i<=n;i++) {
			fact[i] = (fact[i-1] * i) % m;
			ifact[i] = (ifact[i-1] *mi(i)) % m;
		}
		long[] pre = new long[n+1], post = new long[n+1];
		pre[0] = 1;
		post[n] = 1;
		for(int i=1;i<=n;i++) {
			pre[i] = (pre[i-1] * (xi-i+1)+m) % m;
		}
		for(int i=n-1;i>=0;i--) {
			post[i] = (post[i+1] * (xi-i)+m) % m;
		}
		// i-0 * i-1 * i-2 * i-3 * i-4
		for (int i = 0; i < n; i++) {
			long t = (((pre[i] * post[i+1]) % m) * ys[i]) % m;
			t = (((t * ifact[i]) % m) * ifact[n-1-i]) % m;
			if((n-1-i) % 2 == 1)
				t *= -1;
			result += t;
		}
		return ((result % m) + m) % m;
	}

	static long mi(long x) {
		return pow(x, m - 2);
	}

	static long pow(long x, long exp) {
		if (exp < 0) {
			return pow(mi(x), -exp);
		}
		long ans = 1;
		while (exp > 0) {
			if ((exp & 1) == 1)
				ans = (ans * x) % m;
			x = (x * x) % m;
			exp >>= 1;
		}
		return ans;
	}
}