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;
long long dp[5001][5001];
long long n, k;
long long modp(long long a, long long b) {
  if (b == 0) return 1;
  long long u = modp(a, b / 2);
  u = (u * u) % mod;
  if (b & 1) return (a * u) % mod;
  return u;
}
long long f(long long a, long long b) {
  if (dp[a][b] != -1) return dp[a][b];
  long long c = n - b;
  if (a == 0) return dp[a][b] = modp(2, c);
  long long res = 0;
  if (c != 0) {
    res = (c * f(a - 1, b + 1)) % mod;
  }
  if (b != 0) {
    res += (b * f(a - 1, b)) % mod;
    res %= mod;
  }
  return dp[a][b] = res;
}
int32_t main() {
  cin >> n >> k;
  memset(dp, -1, sizeof dp);
  cout << f(k, 0);
}
