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 N = 5e3 + 100;
const int mod = 1e9 + 7;
long long dp[N][N];
long long g[N];
long long PowerMod(long long a, long long b) {
  long long ans = 1;
  a = a % mod;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
void solve() {
  long long n;
  int k;
  scanf("%lld%d", &n, &k);
  long long ans = 0;
  long long temp = n;
  if (n <= k) {
    for (int i = 1; i <= n; i++) {
      ans += (temp * PowerMod(i, k));
      ans %= mod;
      temp = temp * (n - i) % mod * PowerMod(i + 1, mod - 2) % mod;
    }
  } else {
    dp[0][k] = PowerMod(2, n - k);
    for (int i = k - 1; i >= 0; i--) dp[0][i] = dp[0][i + 1] * 2 % mod;
    for (int i = 1; i <= k; i++) {
      for (int j = 0; j <= k; j++) {
        dp[i][j] =
            ((n - j) * (dp[i - 1][j + 1] + g[j + 1]) % mod + j * dp[i - 1][j]) %
            mod;
        g[j + 1] = g[j + 1] * (j + 1) % mod;
      }
    }
    ans = dp[k][0];
  }
  printf("%lld\n", ans);
}
int main() {
  solve();
  return 0;
}
