Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long S[5009][5009], ans, n, k;
const long long mod = 1e9 + 7;
long long read() {
  long long f = 1, re = 0;
  char ch;
  for (ch = getchar(); !isdigit(ch) && ch != '-'; ch = getchar())
    ;
  if (ch == '-') {
    f = -1, ch = getchar();
  }
  for (; isdigit(ch); ch = getchar()) re = (re << 3) + (re << 1) + ch - '0';
  return re * f;
}
long long ksm(long long a, long long b) {
  long long mi = 1;
  while (b) {
    if (b & 1) mi = mi * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return mi;
}
long long P(long long x) {
  long long mi = 1;
  for (long long i = n - x + 1; i <= n; i++) mi = mi * i % mod;
  return mi;
}
signed main() {
  n = read(), k = read();
  S[0][0] = 1;
  for (long long i = 1; i <= k; i++) {
    S[i][0] = 0;
    for (long long j = 1; j <= i; j++)
      S[i][j] = (S[i - 1][j - 1] + j * S[i - 1][j] % mod) % mod;
  }
  for (long long i = 1; i <= min(n, k); i++)
    ans = (ans + S[k][i] * ksm(2, n - i) % mod * P(i) % mod) % mod;
  cout << ans << endl;
  return 0;
}