#include <bits/stdc++.h>
int main() {
  int n, m, i;
  long long ans, now;
  scanf("%d%d", &n, &m);
  ans = 2 * m, now = m;
  for (i = 2; i <= n; i++)
    ans = (ans * (2 * m - 1) + now) % 1000000007, now = now * m % 1000000007;
  printf("%lld\n", ans);
  return 0;
}

### 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>
int main() {
  int n, m, i;
  long long ans, now;
  scanf("%d%d", &n, &m);
  ans = 2 * m, now = m;
  for (i = 2; i <= n; i++)
    ans = (ans * (2 * m - 1) + now) % 1000000007, now = now * m % 1000000007;
  printf("%lld\n", ans);
  return 0;
}

```