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;
inline long long getInt() {
  long long ret;
  char ch;
  bool flg = false;
  while ((ch = getchar()) < '0' || ch > '9') flg |= (ch == '-');
  ret = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') ret = ret * 10 + (ch - '0');
  return flg ? -ret : ret;
}
const int MOD = 1000000007;
inline int advPow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = (long long)ret * a % MOD;
    a = (long long)a * a % MOD;
    b >>= 1;
  }
  return ret;
}
const int MAXK = 5050;
int dp[2][MAXK];
int main() {
  int N = getInt(), K = getInt();
  for (int i = 1; i <= K; i++) {
    dp[i & 1][i] = 1;
    for (int j = 1; j < i; j++)
      dp[i & 1][j] =
          (dp[!(i & 1)][j - 1] + (long long)dp[!(i & 1)][j] * j) % MOD;
  }
  int ans = 0, fc = 1;
  for (int i = 1; i <= min(N, K); i++) {
    fc = (long long)fc * (N - i + 1) % MOD;
    ans = (ans + (long long)dp[K & 1][i] * fc % MOD * advPow(2, N - i)) % MOD;
  }
  printf("%d\n", ans);
}
