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;
template <class T>
inline bool chkmin(T& x, T y) {
  return x > y ? x = y, true : false;
}
template <class T>
inline bool chkmax(T& x, T y) {
  return x < y ? x = y, true : false;
}
inline long long read(void) {
  long long x, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (x = 0; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
inline long long ppow(long long a, long long b) {
  long long ret = 1;
  for (; b; b >>= 1) {
    if (b & 1) ret = ret * a % 1000000007;
    a = a * a % 1000000007;
  }
  return ret;
}
int s[5005][5005], n, k, ans;
int A(int x, int y) {
  int ans = 1;
  for (int i = (1); i <= (y); ++i) ans = 1ll * ans * (x - i + 1) % 1000000007;
  return ans;
}
int main(void) {
  s[0][0] = 1;
  for (int i = (1); i <= (5000); ++i)
    for (int j = (1); j <= (i); ++j)
      s[i][j] = (s[i - 1][j - 1] + 1ll * j * s[i - 1][j]) % 1000000007;
  n = read();
  k = read();
  for (int i = (1); i <= (min(n, k)); ++i)
    (ans += 1ll * s[k][i] * A(n, i) % 1000000007 * ppow(2, n - i) %
            1000000007) %= 1000000007;
  printf("%d\n", ans);
  return 0;
}
