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

```