#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const long long md = (long long)(1e9) + 7;
long long n, m;
long long fpw(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res *= x, res %= md;
    x *= x;
    x %= md;
    y >>= 1;
  }
  return res % md;
}
long long exgcd(long long& x, long long& y, long long a, long long b) {
  if (!b) {
    x = 1;
    y = 0;
    return a;
  }
  long long res = exgcd(y, x, b, a % b);
  y -= x * (a / b);
  return res;
}
long long inv(long long o) {
  long long x, y;
  exgcd(x, y, o, md);
  x = (x % md + md) % md;
  return x;
}
long long ans;
int main() {
  cin >> n >> m;
  long long y;
  for (y = 1; y <= n; ++y) {
    long long tmp = fpw(m, n - y + 1) * fpw(m + m - 1, y - 1);
    tmp %= md;
    ans += tmp;
    ans %= md;
  }
  ans += fpw(m, n);
  ans %= md;
  cout << ans;
}

### Prompt

Please provide a cpp coded solution 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 comment(linker, "/STACK:102400000,102400000")
using namespace std;
const long long md = (long long)(1e9) + 7;
long long n, m;
long long fpw(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res *= x, res %= md;
    x *= x;
    x %= md;
    y >>= 1;
  }
  return res % md;
}
long long exgcd(long long& x, long long& y, long long a, long long b) {
  if (!b) {
    x = 1;
    y = 0;
    return a;
  }
  long long res = exgcd(y, x, b, a % b);
  y -= x * (a / b);
  return res;
}
long long inv(long long o) {
  long long x, y;
  exgcd(x, y, o, md);
  x = (x % md + md) % md;
  return x;
}
long long ans;
int main() {
  cin >> n >> m;
  long long y;
  for (y = 1; y <= n; ++y) {
    long long tmp = fpw(m, n - y + 1) * fpw(m + m - 1, y - 1);
    tmp %= md;
    ans += tmp;
    ans %= md;
  }
  ans += fpw(m, n);
  ans %= md;
  cout << ans;
}

```