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 n, k, ans, mod = 1e9 + 7;
long long s[5010][5010];
long long read() {
  long long r = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') r = (r << 1) + (r << 3) + c - '0', c = getchar();
  return r * f;
}
void put(long long x) {
  if (x < 0) {
    putchar('-');
    x = ~x + 1;
  }
  if (x > 9) put(x / 10);
  putchar(x % 10 + '0');
}
long long poww(long long x, long long y, long long z) {
  long long s = 1;
  while (y) {
    if (y & 1) s = (s * x) % mod;
    x = (x * x) % mod, y >>= 1;
  }
  return s % mod;
}
signed main() {
  n = read(), k = read(), s[0][0] = 1;
  for (long long i = 1; i <= k; i++)
    for (long long j = 1; j <= i; j++)
      s[i][j] = (s[i - 1][j - 1] + (s[i - 1][j] % mod * j) % mod) % mod;
  long long mul = 1, pn = poww(2, n, mod), inv2 = poww(2, mod - 2, mod);
  for (long long i = 0; i <= min(n, k); i++) {
    long long sum = s[k][i] % mod * pn % mod * mul % mod;
    ans = (ans + sum) % mod, pn = (pn * inv2) % mod,
    mul = (mul * (n - i)) % mod;
  }
  put(ans);
  puts("");
  return 0;
}
