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 int kmax = 5e3 + 10;
const int mod = 1e9 + 7;
int n, k;
int dp[kmax][kmax];
int sum(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
int mul(int a, int b) {
  long long c = 1LL * a * b;
  if (c >= mod) c %= mod;
  return c;
}
int lgpow(int a, int b) {
  int res = 1;
  for (int i = 0; (1 << i) <= b; ++i) {
    if (b & (1 << i)) res = mul(res, a);
    a = mul(a, a);
  }
  return res;
}
int get_dp(int a, int b, int c) {
  if (a == 0) dp[a][b] = lgpow(2, c);
  if (dp[a][b] >= 0) return dp[a][b];
  int F = (b == 0) ? 0 : mul(b, get_dp(a - 1, b, c));
  int S = (c == 0) ? 0 : mul(c, get_dp(a - 1, b + 1, c - 1));
  dp[a][b] = sum(F, S);
  return dp[a][b];
}
int main() {
  ios_base ::sync_with_stdio(false);
  for (int i = 0; i < kmax; ++i)
    for (int j = 0; j < kmax; ++j) dp[i][j] = -1;
  cin >> n >> k;
  cout << get_dp(k, 0, n) << '\n';
  return 0;
}
