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 int mod = 1e9 + 7;
long long int pow_mod(long long int x, long long int n) {
  x %= mod;
  long long int res = 1;
  while (n) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
long long int fun(long long int k, long long int a) { long long int res = 1; }
long long int dp[5010][5010];
long long int dfs(long long int k, long long int a, long long int b) {
  if (~dp[k][a]) return dp[k][a];
  if (k == 0) {
    return dp[0][a] = pow_mod(2, b);
  }
  if (b == 0) {
    return dp[k][a] = pow_mod(a, k);
  }
  long long int res = 0;
  res += a * dfs(k - 1, a, b) % mod;
  res += b * dfs(k - 1, a + 1, b - 1) % mod;
  res %= mod;
  return dp[k][a] = res;
}
int main() {
  memset(dp, -1, sizeof(dp));
  long long int n, k;
  scanf("%lld%lld", &n, &k);
  long long int ans = dfs(k, 0, n);
  printf("%lld\n", ans);
}
