Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 5005;
const int mod = 1e9 + 7;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  x *= f;
}
int n, m;
int f[N][N];
int ans;
int Qpow(int base, int ind) {
  int res = 1;
  while (ind) {
    if (ind & 1) res = 1ll * res * base % mod;
    base = 1ll * base * base % mod;
    ind >>= 1;
  }
  return res;
}
int main() {
  read(n), read(m);
  f[0][0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= m; j++)
      f[i][j] = (f[i - 1][j - 1] + 1ll * f[i - 1][j] * j % mod) % mod;
  int fac = 1;
  for (int i = 0; i <= min(n, m); i++) {
    ans += 1ll * f[m][i] * fac % mod * Qpow(2, n - i) % mod;
    ans %= mod;
    fac = 1ll * fac * (n - i) % mod;
  }
  printf("%d\n", ans);
  return 0;
}