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;
long long MOD = 1000000007;
long long N, k, ans, c1, c2;
long long c[5050][5050];
long long a[5050], p[5050], y[5050];
long long pwr(long long a, long long b) {
  long long x[30];
  long long cur = 1;
  if (b % 2 == 1) {
    cur = a;
  }
  x[0] = a;
  for (int i = 1; i < 30; i++) {
    x[i] = x[i - 1] * x[i - 1] % MOD;
    if (b & (1 << i)) {
      cur = cur * x[i] % MOD;
    }
  }
  return cur;
}
long long f1(int a) {
  long long cur = 1;
  for (int i = 0; i <= k; i++) {
    if (i != a) {
      cur = cur * (N - i + MOD) % MOD;
    }
  }
  return cur;
}
long long f2(long long a) {
  long long cur = 1;
  for (long long i = 0; i <= k; i++) {
    if (i != a) {
      cur = cur * (MOD + a - i) % MOD;
    }
  }
  return cur;
}
int main() {
  cin >> N >> k;
  c[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;
    }
  }
  for (int i = 1; i <= k; i++) {
    p[i] = pwr(i, k);
  }
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= i; j++) {
      y[i] += p[j] * c[i][j];
      y[i] %= MOD;
    }
    y[i] = y[i] * pwr(pwr(2, i), MOD - 2) % MOD;
  }
  for (int i = 0; i <= k; i++) {
    c1 = f1(i) * y[i] % MOD;
    c2 = f2(i);
    ans += c1 * pwr(c2, MOD - 2);
    ans %= MOD;
  }
  ans = ans * pwr(2, N) % MOD;
  cout << ans << endl;
}
