#include <bits/stdc++.h>
using namespace std;
long long f[1111111], g[1111111];
void init(int n, int m) {
  f[0] = g[0] = 1;
  for (int i = 1; i <= n; i++)
    f[i] = 1ll * m * f[i - 1] % 1000000007ll,
    g[i] = 1ll * (2 * m - 1) * g[i - 1] % 1000000007ll;
}
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    init(n, m);
    long long ans = 0;
    for (int i = 1; i <= n; i++)
      ans = (ans + f[n - i + 1] * g[i - 1] % 1000000007ll) % 1000000007ll;
    ans = (ans + f[n]) % 1000000007ll;
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Your task is to create 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;
long long f[1111111], g[1111111];
void init(int n, int m) {
  f[0] = g[0] = 1;
  for (int i = 1; i <= n; i++)
    f[i] = 1ll * m * f[i - 1] % 1000000007ll,
    g[i] = 1ll * (2 * m - 1) * g[i - 1] % 1000000007ll;
}
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    init(n, m);
    long long ans = 0;
    for (int i = 1; i <= n; i++)
      ans = (ans + f[n - i + 1] * g[i - 1] % 1000000007ll) % 1000000007ll;
    ans = (ans + f[n]) % 1000000007ll;
    printf("%I64d\n", ans);
  }
  return 0;
}

```