#include <bits/stdc++.h>
using namespace std;
inline string Next() {
  string S;
  cin >> S;
  return S;
}
inline string NextLine() {
  string S;
  getline(cin, S);
  return S;
}
inline int NextInt() {
  int n;
  scanf("%d", &n);
  return n;
}
inline long long NextLong() {
  long long n;
  scanf("%lld", &n);
  return n;
}
inline double NextDouble() {
  double n;
  scanf("%lf", &n);
  return n;
}
const int N = (int)1e6 + 7;
const int P = (int)1e9 + 7;
int n, m, f[N];
int main() {
  n = NextInt();
  m = NextInt();
  long long t = 1;
  for (int i = 1; i <= n; ++i) {
    t = t * m % P;
    f[i] = (f[i - 1] * (2 * m - 1ll) + t) % P;
  }
  cout << (f[n] + t) % P;
  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;
inline string Next() {
  string S;
  cin >> S;
  return S;
}
inline string NextLine() {
  string S;
  getline(cin, S);
  return S;
}
inline int NextInt() {
  int n;
  scanf("%d", &n);
  return n;
}
inline long long NextLong() {
  long long n;
  scanf("%lld", &n);
  return n;
}
inline double NextDouble() {
  double n;
  scanf("%lf", &n);
  return n;
}
const int N = (int)1e6 + 7;
const int P = (int)1e9 + 7;
int n, m, f[N];
int main() {
  n = NextInt();
  m = NextInt();
  long long t = 1;
  for (int i = 1; i <= n; ++i) {
    t = t * m % P;
    f[i] = (f[i - 1] * (2 * m - 1ll) + t) % P;
  }
  cout << (f[n] + t) % P;
  return 0;
}

```