#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e6 + 10;
const long long mod = 1e9 + 7;
const long long inf = 1e9;
long long n, m;
long long fac[maxn];
long long pw(long long a, long long b) {
  if (b == 0) return 1;
  long long k = pw(a, b / 2);
  k = (k * k) % mod;
  if (b & 1) k = (k * a) % mod;
  return k;
}
long long C(long long n, long long k) {
  return (fac[n] * pw((fac[k] * fac[n - k]) % mod, mod - 2)) % mod;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  long long ans = pw(m, n);
  for (long long j = 1; j <= n; j++) {
    (ans += (pw(2 * m - 1, j - 1) * pw(m, n - j + 1)) % mod) %= mod;
  }
  cout << ans;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long maxn = 1e6 + 10;
const long long mod = 1e9 + 7;
const long long inf = 1e9;
long long n, m;
long long fac[maxn];
long long pw(long long a, long long b) {
  if (b == 0) return 1;
  long long k = pw(a, b / 2);
  k = (k * k) % mod;
  if (b & 1) k = (k * a) % mod;
  return k;
}
long long C(long long n, long long k) {
  return (fac[n] * pw((fac[k] * fac[n - k]) % mod, mod - 2)) % mod;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  long long ans = pw(m, n);
  for (long long j = 1; j <= n; j++) {
    (ans += (pw(2 * m - 1, j - 1) * pw(m, n - j + 1)) % mod) %= mod;
  }
  cout << ans;
}

```