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;
char buf[1 << 20], *p1, *p2;
template <class T>
inline void read(T &n) {
  char ch =
      (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2)
           ? 0
           : *p1++);
  T w = 1, x = 0;
  while (!isdigit(ch)) {
    if (ch == '-') w = -1;
    ch =
        (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2)
             ? 0
             : *p1++);
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch =
        (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2)
             ? 0
             : *p1++);
  }
  n = x * w;
}
const int maxn = 1000005;
const int mod = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  return x >= mod ? x - mod : x;
}
inline int mul(int x, int y) { return 1ll * x * y % mod; }
int quick_pow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = mul(res, x);
    x = mul(x, x), y >>= 1;
  }
  return res;
}
int n, k, fac[maxn], inv[maxn];
int cnt[5005];
int main() {
  read(n), read(k);
  cnt[0] = 1;
  int tot = 0, ans = 0;
  for (int i = 1; i <= k; i++) {
    if (n - tot > 0) tot++;
    for (int j = tot; j >= 1; j--)
      cnt[j] = add(mul(cnt[j], n - j), mod - mul(cnt[j - 1], n - j + 1));
    cnt[0] = mul(cnt[0], n);
  }
  for (int i = 0; i <= tot; i++)
    ans = add(ans, mul(cnt[i], quick_pow(2, n - i) - 1));
  cout << ans << endl;
  return 0;
}
