#include <bits/stdc++.h>
using namespace std;
int const mod = 1e9 + 7;
inline int sum(int a, int b) { return (a + b) % mod; }
inline int mul(int a, int b) { return a * 1LL * b % mod; }
int n, m, ans, s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  ans = 1, s = 0;
  for (int i = 1; i <= n; i++) {
    int tmp = ans;
    ans = sum(mul(ans, 2 * m), mul(mod - s, m));
    s = sum(mul(s, m - 1), tmp);
  }
  cout << ans << endl;
}

### 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;
int const mod = 1e9 + 7;
inline int sum(int a, int b) { return (a + b) % mod; }
inline int mul(int a, int b) { return a * 1LL * b % mod; }
int n, m, ans, s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  ans = 1, s = 0;
  for (int i = 1; i <= n; i++) {
    int tmp = ans;
    ans = sum(mul(ans, 2 * m), mul(mod - s, m));
    s = sum(mul(s, m - 1), tmp);
  }
  cout << ans << endl;
}

```