#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
const double PI = acos(-1);
const int oo = 1e9, bpr = 1e9 + 7, N = 1e6 + 100;
int d[N], n, m;
inline int bin_pow(int x, int y) {
  if (!y) return 1;
  if (y & 1) return bin_pow(x, y - 1) * 1ll * x % bpr;
  int c = bin_pow(x, y / 2);
  return c * 1ll * c % bpr;
}
inline void add(int &x, int y) {
  x += y;
  if (x >= bpr) x -= bpr;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  d[0] = 1;
  int ans = bin_pow(m, n);
  int pre = 1;
  for (int i = 1; i <= n; ++i) {
    d[i] = pre * 1ll * m % bpr;
    pre = pre * 1ll * (m - 1) % bpr;
    add(pre, d[i]);
    add(ans, d[i] * 1ll * bin_pow(m, n - i) % bpr);
  }
  cout << 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>
using namespace std;
const double EPS = 1e-9;
const double PI = acos(-1);
const int oo = 1e9, bpr = 1e9 + 7, N = 1e6 + 100;
int d[N], n, m;
inline int bin_pow(int x, int y) {
  if (!y) return 1;
  if (y & 1) return bin_pow(x, y - 1) * 1ll * x % bpr;
  int c = bin_pow(x, y / 2);
  return c * 1ll * c % bpr;
}
inline void add(int &x, int y) {
  x += y;
  if (x >= bpr) x -= bpr;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  d[0] = 1;
  int ans = bin_pow(m, n);
  int pre = 1;
  for (int i = 1; i <= n; ++i) {
    d[i] = pre * 1ll * m % bpr;
    pre = pre * 1ll * (m - 1) % bpr;
    add(pre, d[i]);
    add(ans, d[i] * 1ll * bin_pow(m, n - i) % bpr);
  }
  cout << ans;
  return 0;
}

```