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.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5000;
const int mod = 1e9 + 7;
inline int Add(int x) { return x >= mod ? x - mod : x; }
inline void Add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
inline int Mul(int x, int y) { return (long long)x * y % mod; }
int Pow(int x, int y) {
  int z = 1;
  for (; y; y >>= 1) {
    if (y & 1) z = Mul(z, x);
    x = Mul(x, x);
  }
  return z;
}
int inv[maxn + 1];
int Choose(int n, int m) {
  int ans = 1;
  for (int i = 1; i <= m; ++i) {
    ans = Mul(ans, n - i + 1);
    ans = Mul(ans, inv[i]);
  }
  return ans;
}
int n, k;
int f[maxn + 1][maxn + 1];
int main(void) {
  f[0][0] = 1;
  for (int i = 1; i <= maxn; ++i) {
    for (int j = 1; j <= i; ++j) {
      f[i][j] = Mul(Add(f[i - 1][j] + f[i - 1][j - 1]), j);
    }
  }
  for (int i = 1; i <= maxn; ++i) {
    inv[i] = Pow(i, mod - 2);
  }
  scanf("%d%d", &n, &k);
  int ans = 0;
  for (int t = 1; t <= min(n, k); ++t) {
    Add(ans, Mul(Choose(n, t), Mul(f[k][t], Pow(2, n - t))));
  }
  printf("%d\n", ans);
  return 0;
}
