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

### Prompt

Construct a CPP code solution to the problem outlined:

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 long long p = 1e9 + 7;
long long n, m, i, j, ans, f[1000010], jc[1000010];
int main() {
  scanf("%lld%lld", &n, &m);
  jc[0] = 1;
  for (i = 1; i <= n; i++) jc[i] = jc[i - 1] * m % p;
  f[0] = 1;
  for (i = 1; i <= n; i++) {
    ans = (ans + jc[n - i] * f[i - 1] % p * m % p) % p;
    f[i] = f[i - 1] * (m + m - 1) % p;
  }
  printf("%lld", (ans + jc[n]) % p);
}

```