#include <bits/stdc++.h>
using namespace std;
const long long p = 1000000007LL;
long long power(long long x, long long n) {
  long long ret = 1LL;
  while (n > 0LL) {
    if (n & 1LL) ret = ret * x % p;
    x = x * x % p;
    n >>= 1LL;
  }
  return ret;
}
long long n, m;
long long dp;
int main() {
  cin >> n >> m;
  dp = 1;
  for (long long i = 1LL; i <= n; i++)
    dp = (dp * (m * 2LL - 1LL) + power(m, i - 1LL)) % p;
  cout << dp << endl;
}

### Prompt

Create a solution in Cpp for 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 long long p = 1000000007LL;
long long power(long long x, long long n) {
  long long ret = 1LL;
  while (n > 0LL) {
    if (n & 1LL) ret = ret * x % p;
    x = x * x % p;
    n >>= 1LL;
  }
  return ret;
}
long long n, m;
long long dp;
int main() {
  cin >> n >> m;
  dp = 1;
  for (long long i = 1LL; i <= n; i++)
    dp = (dp * (m * 2LL - 1LL) + power(m, i - 1LL)) % p;
  cout << dp << endl;
}

```