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 long long Mod = 1e9 + 7;
const long long M = 5e3 + 10;
long long n, ans, k, s[M][M], dfac[M];
long long poww(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) r = r * a % Mod;
    a = a * a % Mod;
    b >>= 1;
  }
  return r;
}
long long fac[M], inv[M];
void init() {
  fac[0] = 1;
  for (long long i = 1; i < M; i++) fac[i] = fac[i - 1] * i % Mod;
  for (long long i = 0; i < M; i++) inv[i] = poww(fac[i], Mod - 2);
}
long long p = Mod;
long long C(long long m, long long n) {
  return fac[m] * inv[m - n] % p * inv[n] % p;
}
signed main() {
  cin >> n >> k;
  s[0][0] = 1;
  if (k >= n) {
    init();
    for (long long i = 1; i <= n; i++) ans = (ans + C(n, i) * poww(i, k)) % Mod;
    cout << ans;
    return 0;
  }
  for (long long i = 1; i <= k; i++) {
    for (long long j = 1; j <= i; j++)
      s[i][j] = (s[i - 1][j - 1] + j * s[i - 1][j] % Mod) % Mod;
  }
  dfac[0] = 1;
  for (long long i = 1; i <= k; i++) dfac[i] = dfac[i - 1] * (n - i + 1) % Mod;
  for (long long i = 0; i <= k; i++)
    ans = (ans + s[k][i] * dfac[i] % Mod * poww(2, n - i) % Mod) % Mod;
  cout << ans;
  return 0;
}
