#include <bits/stdc++.h>
const int MAX_N = (int)1e6 + 123;
const double eps = 1e-6;
const int mod = (int)1e9 + 7;
using namespace std;
int n, m;
int binpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return res;
}
int first[MAX_N], _f[MAX_N];
int C(int n, int k) {
  int res = 1ll * first[n] * _f[k] % mod;
  res = 1ll * res * _f[n - k] % mod;
  return res;
}
int main() {
  cin >> n >> m;
  first[0] = _f[0] = 1;
  for (int i = 1; i <= (int)1e6; i++) {
    first[i] = 1ll * first[i - 1] * i % mod;
    _f[i] = binpow(first[i], mod - 2);
  }
  int res = binpow(m, n);
  for (int i = 1; i <= n; i++) {
    int now = binpow(m, i) * 1ll * binpow(m - 1, n - i) % mod;
    now = 1ll * now * C(n, i - 1) % mod;
    res = (res + now) % mod;
  }
  cout << res;
  return 0;
}

### 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>
const int MAX_N = (int)1e6 + 123;
const double eps = 1e-6;
const int mod = (int)1e9 + 7;
using namespace std;
int n, m;
int binpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return res;
}
int first[MAX_N], _f[MAX_N];
int C(int n, int k) {
  int res = 1ll * first[n] * _f[k] % mod;
  res = 1ll * res * _f[n - k] % mod;
  return res;
}
int main() {
  cin >> n >> m;
  first[0] = _f[0] = 1;
  for (int i = 1; i <= (int)1e6; i++) {
    first[i] = 1ll * first[i - 1] * i % mod;
    _f[i] = binpow(first[i], mod - 2);
  }
  int res = binpow(m, n);
  for (int i = 1; i <= n; i++) {
    int now = binpow(m, i) * 1ll * binpow(m - 1, n - i) % mod;
    now = 1ll * now * C(n, i - 1) % mod;
    res = (res + now) % mod;
  }
  cout << res;
  return 0;
}

```