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;
void read(int &x) {
  char ch;
  bool ok;
  for (ok = 0, ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') ok = 1;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  if (ok) x = -x;
}
const int maxn = 5010, N = 5000, mod = 1e9 + 7;
int n, k, ans, fac[maxn], inv[maxn], s[maxn][maxn];
int mul(int x, int y) { return 1ll * x * y - 1ll * x * y / mod * mod; }
int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
int del(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
int mi(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    b >>= 1, a = mul(a, a);
  }
  return ans;
}
int C(int n, int j) {
  int ans = 1;
  for (register int i = n; i >= n - j + 1; i--) ans = mul(ans, i);
  return mul(ans, mi(fac[j], mod - 2));
}
int main() {
  read(n), read(k);
  s[0][0] = fac[0] = 1;
  for (register int i = 1; i <= k; i++)
    for (register int j = 1; j <= i; j++)
      s[i][j] = add(s[i - 1][j - 1], mul(j, s[i - 1][j]));
  for (register int i = 1; i <= k; i++) fac[i] = mul(fac[i - 1], i);
  for (register int j = 0; j <= min(k, n); j++)
    ans = add(ans, mul(s[k][j], mul(fac[j], mul(C(n, j), mi(2, n - j)))));
  printf("%d\n", ans);
}