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>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/STACK:16777216")
using namespace std;
using matrix = vector<vector<long long>>;
void FAST_IO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long mod = 1e9 + 7;
const int MAXN = 5050;
long long dp[MAXN][MAXN];
long long binpow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res = (res * a) % mod;
    }
    a = (a * a) % mod;
    n >>= 1;
  }
  return res;
}
signed main() {
  FAST_IO();
  int n, k;
  cin >> n >> k;
  dp[0][0] = 1;
  for (int i = 1; i < MAXN; ++i) {
    for (int j = 1; j <= i; ++j) {
      dp[i][j] = (dp[i - 1][j] * j + dp[i - 1][j - 1] * (n - j + 1)) % mod;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= min(n, k); ++i) {
    ans = (ans + dp[k][i] * binpow(2, n - i)) % mod;
  }
  cout << ans;
}
