#include <bits/stdc++.h>
using namespace std;
const long long base = 1000000007;
long long n, m;
long long power(long long a, int n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * a % base;
    a = a * a % base;
    n >>= 1;
  }
  return res;
}
int main() {
  cin >> n >> m;
  if (m == 1) {
    cout << (n + 1) % base;
    return 0;
  }
  long long res = power(m, n) * 2;
  res = (res +
         (m * power(2 * m - 1, n) - (2 * m - 1) * power(m, n) + base * base) %
             base * power(m - 1, base - 2) % base) %
        base;
  cout << res;
}

### 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 base = 1000000007;
long long n, m;
long long power(long long a, int n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * a % base;
    a = a * a % base;
    n >>= 1;
  }
  return res;
}
int main() {
  cin >> n >> m;
  if (m == 1) {
    cout << (n + 1) % base;
    return 0;
  }
  long long res = power(m, n) * 2;
  res = (res +
         (m * power(2 * m - 1, n) - (2 * m - 1) * power(m, n) + base * base) %
             base * power(m - 1, base - 2) % base) %
        base;
  cout << res;
}

```