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 MAXN = 5005;
const long long MOD = 1e9 + 7;
template <typename T>
inline void read(T &AKNOI) {
  T x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  AKNOI = flag * x;
}
int n, k, mn, stir[MAXN][MAXN];
inline long long ksm(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % MOD;
    x = x * x % MOD;
    y >>= 1;
  }
  return ret;
}
void init() {
  read(n);
  read(k);
  mn = min(n, k);
  stir[0][0] = 1;
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= k; ++j) {
      stir[i][j] = (stir[i - 1][j - 1] + 1LL * j * stir[i - 1][j]) % MOD;
    }
  }
}
void solve() {
  long long ans = 0, fac = 1, m2 = ksm(2, n), inv2 = ksm(2, MOD - 2);
  for (int i = 0; i <= mn; ++i) {
    ans = (ans + fac * m2 % MOD * stir[k][i] % MOD) % MOD;
    m2 = m2 * inv2 % MOD;
    fac = fac * (n - i) % MOD;
  }
  printf("%lld\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}
