#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000000 + 3;
const int MOD = (int)1e9 + 7;
long long p1[MAXN], p2[MAXN];
void init(int n, int m) {
  p1[0] = 1, p2[0] = 1;
  for (int i = 1; i <= n; i++) {
    p1[i] = p1[i - 1] * m % MOD;
    p2[i] = p2[i - 1] * (2 * m - 1) % MOD;
  }
}
long long solve(int n, int m) {
  init(n, m);
  long long ans = p1[n];
  for (int i = 0; i < n; i++) {
    ans = (ans + p1[n - i] * p2[i] % MOD) % MOD;
  }
  return ans;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  printf("%d\n", (int)solve(n, m));
  return 0;
}

### 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 int MAXN = 1000000 + 3;
const int MOD = (int)1e9 + 7;
long long p1[MAXN], p2[MAXN];
void init(int n, int m) {
  p1[0] = 1, p2[0] = 1;
  for (int i = 1; i <= n; i++) {
    p1[i] = p1[i - 1] * m % MOD;
    p2[i] = p2[i - 1] * (2 * m - 1) % MOD;
  }
}
long long solve(int n, int m) {
  init(n, m);
  long long ans = p1[n];
  for (int i = 0; i < n; i++) {
    ans = (ans + p1[n - i] * p2[i] % MOD) % MOD;
  }
  return ans;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  printf("%d\n", (int)solve(n, m));
  return 0;
}

```