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;
template <typename T>
inline void upmin(T &x, T y) {
  y < x ? x = y : 0;
}
template <typename T>
inline void upmax(T &x, T y) {
  x < y ? x = y : 0;
}
const long double pi = acos(-1);
const int oo = 1 << 30;
const long long OO = 1e18;
const int N = 5010;
int gi() {
  int w = 0;
  bool q = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') q = 0, c = getchar();
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return q ? w : -w;
}
int s[N][N];
const int mod = 1e9 + 7;
inline int qpow(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = 1LL * ans * x % mod;
    x = 1LL * x * x % mod;
    y >>= 1;
  }
  return ans;
}
int main() {
  int n = gi(), k = gi(), i, j;
  for (i = s[0][0] = 1; i <= k; i++)
    for (j = 1; j <= i; j++)
      s[i][j] = (1LL * s[i - 1][j] * j + s[i - 1][j - 1]) % mod;
  int ans = 0, fac = 1;
  for (i = 1; i <= k && i <= n; i++) {
    fac = 1LL * fac * (n + 1 - i) % mod;
    ans = (ans + 1LL * s[k][i] * qpow(2, n - i) % mod * fac) % mod;
  }
  cout << ans << endl;
  return 0;
}
