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 = 5005;
const int MOD = 1e9 + 7;
long long f[MAXN][MAXN];
inline int read() {
  int x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
inline int qpow(int x, int y = MOD - 2) {
  long long r = 1, k = x;
  for (; y; y >>= 1) {
    if (y & 1) r *= k, r %= MOD;
    k *= k, k %= MOD;
  }
  return r;
}
int main() {
  int n = read(), k = read();
  f[0][0] = 1;
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= k; j++)
      f[i][j] = (f[i - 1][j - 1] + j * f[i - 1][j]) % MOD;
  long long ans = 0, tmp = 1;
  for (int i = 0; i <= min(k, n); i++) {
    ans = (ans + f[k][i] * tmp % MOD * qpow(2, n - i)) % MOD;
    (tmp *= n - i) %= MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
