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;
using ll = long long;
constexpr int K = 5005;
constexpr ll mod = 1e9 + 7;
ll n, k;
ll S[K][K];
void pre() {
  S[0][0] = 1;
  for (int i = 1; i < K; i++) {
    for (int j = 1; j <= i; j++) {
      S[i][j] = (S[i - 1][j - 1] + 1ll * S[i - 1][j] * j % mod) % mod;
    }
  }
}
ll qpow(ll a, ll b) {
  b %= mod - 1;
  if (b < 0) b += mod - 1;
  ll res = 1;
  while (b) {
    if (b & 1) {
      res = res * a % mod;
    }
    a = a * a % mod;
    b /= 2;
  }
  return res;
}
int main() {
  cin >> n >> k;
  pre();
  auto p2 = qpow(2, n);
  auto inv2 = qpow(2, -1);
  ll ans = 0;
  ll frac = 1;
  for (int i = 0; i <= k; i++) {
    ans = (ans + 1ll * frac * S[k][i] % mod * p2 % mod) % mod;
    frac = frac * (n - i) % mod;
    p2 = p2 * inv2 % mod;
  }
  cout << ans << endl;
  return 0;
}
