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 long long maxn = 5e3 + 10;
const long long mod = 1e9 + 7;
long long n, m, k, ans, a, b;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
long long c[maxn][maxn], f[maxn];
inline long long ksm(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
signed main() {
  n = read(), k = read();
  a = 1, b = ksm(a + 1, mod - 2);
  c[0][0] = 1;
  f[0] = ksm(a + 1, n);
  for (long long i = 1; i <= k; i++) {
    c[i][0] = 1;
    for (long long j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  }
  for (long long i = 1; i <= k; i++) {
    long long t1 = 0, t2 = 0;
    for (long long j = 0; j <= i - 1; j++)
      t1 = (t1 + c[i - 1][j] * f[j] % mod) % mod;
    for (long long j = 0; j <= i - 2; j++)
      t2 = (t2 + c[i - 1][j] * f[j + 1] % mod) % mod;
    f[i] = a * b % mod * (n * t1 % mod - t2 + mod) % mod;
  }
  printf("%lld\n", f[k]);
  return 0;
}
