#include <bits/stdc++.h>
const int MAXN = 1e6 + 5;
const int ha = 1e9 + 7;
inline int qpow(int a, int n = ha - 2) {
  int res = 1;
  while (n) {
    if (n & 1) res = 1ll * res * a % ha;
    a = 1ll * a * a % ha;
    n >>= 1;
  }
  return res;
}
int n, m;
int f[MAXN];
int main() {
  scanf("%d%d", &n, &m);
  f[0] = 1;
  f[1] = 3 * m - 1;
  for (int i = 2; i <= n; ++i) {
    f[i] = 1ll * (3 * m - 1) * f[i - 1] % ha;
    (f[i] += ha - 1ll * m * (2 * m - 1) % ha * f[i - 2] % ha) %= ha;
  }
  int ans = f[n];
  (ans += ha - qpow(m, n)) %= ha;
  ans = 1ll * ans * m % ha * qpow(2 * m - 1) % ha;
  (ans += qpow(m, n)) %= ha;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Develop a solution in CPP 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>
const int MAXN = 1e6 + 5;
const int ha = 1e9 + 7;
inline int qpow(int a, int n = ha - 2) {
  int res = 1;
  while (n) {
    if (n & 1) res = 1ll * res * a % ha;
    a = 1ll * a * a % ha;
    n >>= 1;
  }
  return res;
}
int n, m;
int f[MAXN];
int main() {
  scanf("%d%d", &n, &m);
  f[0] = 1;
  f[1] = 3 * m - 1;
  for (int i = 2; i <= n; ++i) {
    f[i] = 1ll * (3 * m - 1) * f[i - 1] % ha;
    (f[i] += ha - 1ll * m * (2 * m - 1) % ha * f[i - 2] % ha) %= ha;
  }
  int ans = f[n];
  (ans += ha - qpow(m, n)) %= ha;
  ans = 1ll * ans * m % ha * qpow(2 * m - 1) % ha;
  (ans += qpow(m, n)) %= ha;
  printf("%d\n", ans);
  return 0;
}

```