#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
const long long N = 1000010, Mod = 1e9 + 7;
inline long long read() {
  long long s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * w;
}
long long n, m, pw[N + 5];
signed main() {
  n = read(), m = read();
  pw[0] = 1;
  for (register long long i = 1; i <= n; i++) pw[i] = 1ll * pw[i - 1] * m % Mod;
  long long ans = pw[n];
  for (register long long i = 1, now = 1; i <= n; i++) {
    ans = (ans + 1ll * pw[n - i + 1] * now % Mod) % Mod;
    now = 1ll * now * (m * 2 - 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>
#pragma GCC optimize(3)
using namespace std;
const long long N = 1000010, Mod = 1e9 + 7;
inline long long read() {
  long long s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * w;
}
long long n, m, pw[N + 5];
signed main() {
  n = read(), m = read();
  pw[0] = 1;
  for (register long long i = 1; i <= n; i++) pw[i] = 1ll * pw[i - 1] * m % Mod;
  long long ans = pw[n];
  for (register long long i = 1, now = 1; i <= n; i++) {
    ans = (ans + 1ll * pw[n - i + 1] * now % Mod) % Mod;
    now = 1ll * now * (m * 2 - 1) % Mod;
  }
  printf("%d\n", ans);
  return 0;
}

```