#include <bits/stdc++.h>
using namespace std;
long long dp[(int)(1e6 + 10)], last, n, m;
int main() {
  cin >> n >> m;
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    dp[i] = (((dp[i - 1] * m) % 1000000007) * 2) % 1000000007;
    dp[i] = (dp[i] - (last * m) % 1000000007 + 1000000007) % 1000000007;
    last = (dp[i - 1] + (last * (m - 1)) % 1000000007) % 1000000007;
  }
  cout << dp[n] << '\n';
}

### Prompt

In cpp, your task is to solve 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;
long long dp[(int)(1e6 + 10)], last, n, m;
int main() {
  cin >> n >> m;
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    dp[i] = (((dp[i - 1] * m) % 1000000007) * 2) % 1000000007;
    dp[i] = (dp[i] - (last * m) % 1000000007 + 1000000007) % 1000000007;
    last = (dp[i - 1] + (last * (m - 1)) % 1000000007) % 1000000007;
  }
  cout << dp[n] << '\n';
}

```