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;
long long n, m, ans;
long long f[5005], s[5005][5005], g[5005];
inline long long POW(long long a, long long b = 1000000007 - 2,
                     long long ans = 1) {
  for (; b; b /= 2, a = a * a % 1000000007)
    if (b & 1) ans = ans * a % 1000000007;
  return ans;
}
void init(long long n) {
  s[0][0] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= n; j++)
      s[i][j] = (s[i - 1][j - 1] + j * s[i - 1][j] % 1000000007) % 1000000007;
}
signed main() {
  cin >> n >> m;
  init(m);
  f[0] = 1;
  for (long long i = 1; i <= m; i++) f[i] = f[i - 1] * (n - i + 1) % 1000000007;
  for (long long i = 0; i <= m; i++)
    (ans += s[m][i] * f[i] % 1000000007 * POW(2, n - i) % 1000000007) %=
        1000000007;
  cout << ans;
}
