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 mod = 1e9 + 7, N = 1e6 + 1;
int jc[N];
int ijc[N];
int qpow(int a, int b) {
  int ans = 1;
  if (b < 0) a = qpow(a, mod - 2), b = -b;
  while (b) {
    if (b & 1) ans = (long long)ans * a % mod;
    b >>= 1;
    a = (long long)a * a % mod;
  }
  return ans;
}
int C(int n, int m) {
  return (long long)jc[n] * ijc[n - m] % mod * ijc[m] % mod;
}
int n, k;
int upd(int x) {
  while (x >= mod) x -= mod;
  while (x < 0) x += mod;
  return x;
}
void subtask1() {
  int ans = 0;
  for (int i = (1); i <= (n); ++i)
    ans = upd(ans + (long long)C(n, i) * qpow(i, k) % mod);
  printf("%d", ans);
}
int xs[100001];
int ne[N];
int main() {
  scanf("%d%d", &n, &k);
  xs[0] = 1;
  for (int qwq = (1); qwq <= (k); ++qwq) {
    for (int i = (0); i <= (k); ++i) ne[i] = 0;
    for (int i = (0); i <= (k); ++i) {
      ne[i] = upd(ne[i] + (long long)(n - i) * xs[i] % mod);
      ne[i + 1] = upd(ne[i + 1] - (long long)(n - i) * xs[i] % mod);
    }
    for (int i = (0); i <= (k); ++i) xs[i] = ne[i];
  }
  int ans = 0;
  for (int i = (0); i <= (k); ++i)
    ans = upd(ans + (long long)qpow(2, n - i) * xs[i] % mod);
  printf("%d", ans);
}
