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 long long N = 5010;
const long long M = 1e9;
const long long mod = 1e9 + 7;
long long a, b, s[N][N], ans;
inline long long S(long long n, long long m) {
  if (n < m || m == 0) return 0;
  if (n == m || m == 1) return 1;
  if (s[n][m]) return s[n][m];
  return (S(n - 1, m - 1) + m * S(n - 1, m)) % mod;
}
inline long long power(long long n, long long m) {
  long long sum = 1;
  while (m) {
    if (m & 1) sum = sum * n % mod;
    n = n * n % mod;
    m >>= 1;
  }
  return sum;
}
signed main() {
  scanf("%lld%lld", &a, &b);
  for (long long i = 1; i <= b; i++)
    for (long long j = 1; j <= b; j++) s[i][j] = S(i, j);
  long long p1 = 1, p2 = power(2, a), d2 = power(2, mod - 2);
  for (long long i = 0; i <= b; i++) {
    ans = (ans + s[b][i] * p1 % mod * p2 % mod) % mod;
    p2 = 1ll * p2 * d2 % mod;
    p1 = 1ll * p1 * (a - i) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
