#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
long long pm0[1000003], pm1[1000003];
int main() {
  long long n, m;
  scanf("%lld %lld", &n, &m);
  pm0[0] = pm1[0] = 1;
  for (int i = 1; i <= 1000000; i++) {
    pm0[i] = (pm0[i - 1] * m) % MOD;
    pm1[i] = (pm1[i - 1] * (2 * m - 1)) % MOD;
  }
  long long res = pm0[n];
  for (int i = 0; i <= n - 1; i++) {
    res += (pm0[n - i] * pm1[i]) % MOD;
    res %= MOD;
  }
  printf("%lld\n", res);
}

### 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>
using namespace std;
const long long MOD = 1e9 + 7;
long long pm0[1000003], pm1[1000003];
int main() {
  long long n, m;
  scanf("%lld %lld", &n, &m);
  pm0[0] = pm1[0] = 1;
  for (int i = 1; i <= 1000000; i++) {
    pm0[i] = (pm0[i - 1] * m) % MOD;
    pm1[i] = (pm1[i - 1] * (2 * m - 1)) % MOD;
  }
  long long res = pm0[n];
  for (int i = 0; i <= n - 1; i++) {
    res += (pm0[n - i] * pm1[i]) % MOD;
    res %= MOD;
  }
  printf("%lld\n", res);
}

```