#include <bits/stdc++.h>
using namespace std;
long long jc[1000001], Jc[1000001], w[1000001], p[1000001], P[1000001], ans;
int n, m, i, j, k, l;
int main() {
  scanf("%d%d", &n, &m);
  if (m == 1) {
    printf("%d\n", n + 1);
    return 0;
  }
  jc[0] = jc[1] = Jc[0] = Jc[1] = w[1] = p[0] = P[0] = 1;
  p[1] = m + m - 1;
  P[1] = m;
  for (i = 2; i <= n; i++)
    w[i] = 1000000007 - w[1000000007 % i] * (1000000007 / i) % 1000000007,
    jc[i] = jc[i - 1] * i % 1000000007, Jc[i] = Jc[i - 1] * w[i] % 1000000007,
    p[i] = p[i - 1] * (m + m - 1) % 1000000007,
    P[i] = P[i - 1] * m % 1000000007;
  for (i = 1; i <= n; i++) ans = (ans + p[i - 1] * P[n - i + 1]) % 1000000007;
  printf("%lld\n", (ans + P[n]) % 1000000007);
}

### 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 jc[1000001], Jc[1000001], w[1000001], p[1000001], P[1000001], ans;
int n, m, i, j, k, l;
int main() {
  scanf("%d%d", &n, &m);
  if (m == 1) {
    printf("%d\n", n + 1);
    return 0;
  }
  jc[0] = jc[1] = Jc[0] = Jc[1] = w[1] = p[0] = P[0] = 1;
  p[1] = m + m - 1;
  P[1] = m;
  for (i = 2; i <= n; i++)
    w[i] = 1000000007 - w[1000000007 % i] * (1000000007 / i) % 1000000007,
    jc[i] = jc[i - 1] * i % 1000000007, Jc[i] = Jc[i - 1] * w[i] % 1000000007,
    p[i] = p[i - 1] * (m + m - 1) % 1000000007,
    P[i] = P[i - 1] * m % 1000000007;
  for (i = 1; i <= n; i++) ans = (ans + p[i - 1] * P[n - i + 1]) % 1000000007;
  printf("%lld\n", (ans + P[n]) % 1000000007);
}

```