#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
long long dp[2], pw;
int main() {
  int n, m;
  scanf("%d", &n);
  scanf("%d", &m);
  dp[0] = 1;
  pw = 1;
  for (long long i = 1; i <= n; i++) {
    dp[i & 1] = (dp[(i - 1) & 1] * (2 * m - 1) + pw) % MOD;
    pw = (pw * (long long)m) % MOD;
  }
  printf("%d\n", (int)dp[n & 1]);
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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>
using namespace std;
const int MOD = 1e9 + 7;
long long dp[2], pw;
int main() {
  int n, m;
  scanf("%d", &n);
  scanf("%d", &m);
  dp[0] = 1;
  pw = 1;
  for (long long i = 1; i <= n; i++) {
    dp[i & 1] = (dp[(i - 1) & 1] * (2 * m - 1) + pw) % MOD;
    pw = (pw * (long long)m) % MOD;
  }
  printf("%d\n", (int)dp[n & 1]);
  return 0;
}

```