#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e6 + 10, mod = 1e9 + 7;
int n, m;
inline void update(int &x, int y) { x = x + y >= mod ? x + y - mod : x + y; }
inline int mpow(int a, int x) {
  int ans = 1;
  while (x) {
    if (x & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    x >>= 1;
  }
  return ans;
}
int main() {
  scanf("%d %d", &n, &m);
  int ans = mpow(m, n);
  for (int i = 1; i <= n; i++)
    update(ans, 1ll * mpow(m, n - i + 1) * mpow(2 * m - 1, i - 1) % mod);
  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>
using namespace std;
const int maxN = 1e6 + 10, mod = 1e9 + 7;
int n, m;
inline void update(int &x, int y) { x = x + y >= mod ? x + y - mod : x + y; }
inline int mpow(int a, int x) {
  int ans = 1;
  while (x) {
    if (x & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    x >>= 1;
  }
  return ans;
}
int main() {
  scanf("%d %d", &n, &m);
  int ans = mpow(m, n);
  for (int i = 1; i <= n; i++)
    update(ans, 1ll * mpow(m, n - i + 1) * mpow(2 * m - 1, i - 1) % mod);
  printf("%d\n", ans);
  return 0;
}

```