#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
inline int ad(int x, int y) { return (x + y) % mod; }
inline int mu(int x, int y) { return 1LL * x * y % mod; }
int main() {
  int n, m;
  cin >> n >> m;
  int a = 1, b = 0;
  for (int i = 0; i < n; i++) {
    int A, B;
    A = ad(mu(2 * m, a), mu(mod - m, b));
    B = ad(mu(m - 1, b), a);
    a = A, b = B;
  }
  cout << a << endl;
}

### Prompt

Generate 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;
const int mod = 1e9 + 7;
inline int ad(int x, int y) { return (x + y) % mod; }
inline int mu(int x, int y) { return 1LL * x * y % mod; }
int main() {
  int n, m;
  cin >> n >> m;
  int a = 1, b = 0;
  for (int i = 0; i < n; i++) {
    int A, B;
    A = ad(mu(2 * m, a), mu(mod - m, b));
    B = ad(mu(m - 1, b), a);
    a = A, b = B;
  }
  cout << a << endl;
}

```