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>
const double pi = acos(-1);
const long long inf = 1e9 + 7;
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f * x;
}
int n, m;
long long d[140005], ans;
long long qpow(long long x, long long t) {
  long long s = 1;
  for (; t; t >>= 1, x = ((x) * (x)) % inf)
    if (t & 1) s = s * x % inf;
  return s;
}
int main() {
  n = read();
  m = read();
  d[1] = 1;
  for (int i = 2; i <= m; i++)
    for (int j = i; j >= 1; j--) d[j] = (j * d[j] + d[j - 1]) % inf;
  long long t = 1;
  m = min(n, m);
  for (int i = 1; i <= m; i++) {
    t = t * (n - i + 1) % inf;
    ans += d[i] * qpow(2, n - i) % inf * t % inf;
    ans %= inf;
  }
  return 0 * printf("%lld\n", ans);
}
