#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long n, m;
long long pm[1000005], pmm[1000005];
long long ans;
int main() {
  scanf("%I64d%I64d", &n, &m);
  pm[0] = 1;
  for (long long k = 1; k <= n; k++) pm[k] = (pm[k - 1] * m) % mod;
  pmm[0] = 1;
  for (long long k = 1; k <= n; k++) pmm[k] = (pmm[k - 1] * (2 * m - 1)) % mod;
  ans = 0;
  for (long long k = 1; k <= n; k++)
    ans = (ans + pm[n - k + 1] * pmm[k - 1]) % mod;
  ans = (ans + pm[n]) % mod;
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

In CPP, your task is to solve 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 mod = 1000000007;
long long n, m;
long long pm[1000005], pmm[1000005];
long long ans;
int main() {
  scanf("%I64d%I64d", &n, &m);
  pm[0] = 1;
  for (long long k = 1; k <= n; k++) pm[k] = (pm[k - 1] * m) % mod;
  pmm[0] = 1;
  for (long long k = 1; k <= n; k++) pmm[k] = (pmm[k - 1] * (2 * m - 1)) % mod;
  ans = 0;
  for (long long k = 1; k <= n; k++)
    ans = (ans + pm[n - k + 1] * pmm[k - 1]) % mod;
  ans = (ans + pm[n]) % mod;
  printf("%I64d\n", ans);
  return 0;
}

```