#include <bits/stdc++.h>
long long f[1000010], sum[1000010], n, m;
int main() {
  scanf("%lld%lld", &n, &m);
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    f[i] = m * ((int)(1e9 + 7) + f[i - 1] * 2 - sum[i - 1]) % (int)(1e9 + 7);
    sum[i] = (sum[i - 1] * (m - 1) + f[i - 1]) % (int)(1e9 + 7);
  }
  printf("%lld", f[n]);
}

### Prompt

Please provide a CPP coded solution to the problem described below:

For a sequence a of n integers between 1 and m, inclusive, denote f(a) as the number of distinct subsequences of a (including the empty subsequence).

You are given two positive integers n and m. Let S be the set of all sequences of length n consisting of numbers from 1 to m. Compute the sum f(a) over all a in S modulo 109 + 7.

Input

The only line contains two integers n and m (1 ≤ n, m ≤ 106) — the number of elements in arrays and the upper bound for elements.

Output

Print the only integer c — the desired sum modulo 109 + 7.

Examples

Input

1 3


Output

6


Input

2 2


Output

14


Input

3 3


Output

174

### Response

```cpp
#include <bits/stdc++.h>
long long f[1000010], sum[1000010], n, m;
int main() {
  scanf("%lld%lld", &n, &m);
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    f[i] = m * ((int)(1e9 + 7) + f[i - 1] * 2 - sum[i - 1]) % (int)(1e9 + 7);
    sum[i] = (sum[i - 1] * (m - 1) + f[i - 1]) % (int)(1e9 + 7);
  }
  printf("%lld", f[n]);
}

```