#include <bits/stdc++.h>
const int MAXN = 1e6 + 5;
const int ha = 1e9 + 7;
inline int qpow(int a, int n = ha - 2) {
  int res = 1;
  while (n) {
    if (n & 1) res = 1ll * res * a % ha;
    a = 1ll * a * a % ha;
    n >>= 1;
  }
  return res;
}
int fac[MAXN], inv[MAXN];
inline int C(int n, int m) {
  return n < 0 || m < 0 || n < m ? 0
                                 : 1ll * fac[n] * inv[m] % ha * inv[n - m] % ha;
}
int main() {
  fac[0] = 1;
  for (int i = 1; i <= MAXN - 1; ++i) fac[i] = 1ll * fac[i - 1] * i % ha;
  inv[MAXN - 1] = qpow(fac[MAXN - 1]);
  for (int i = MAXN - 2; i >= 0; --i) inv[i] = 1ll * inv[i + 1] * (i + 1) % ha;
  int n, m;
  scanf("%d%d", &n, &m);
  int ans = qpow(m, n);
  for (int i = 1; i <= n; ++i) {
    int sm = 0;
    sm = 1ll * qpow(m, n - i + 1) * qpow(2 * m - 1, i - 1) % ha;
    (ans += sm) %= ha;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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 MAXN = 1e6 + 5;
const int ha = 1e9 + 7;
inline int qpow(int a, int n = ha - 2) {
  int res = 1;
  while (n) {
    if (n & 1) res = 1ll * res * a % ha;
    a = 1ll * a * a % ha;
    n >>= 1;
  }
  return res;
}
int fac[MAXN], inv[MAXN];
inline int C(int n, int m) {
  return n < 0 || m < 0 || n < m ? 0
                                 : 1ll * fac[n] * inv[m] % ha * inv[n - m] % ha;
}
int main() {
  fac[0] = 1;
  for (int i = 1; i <= MAXN - 1; ++i) fac[i] = 1ll * fac[i - 1] * i % ha;
  inv[MAXN - 1] = qpow(fac[MAXN - 1]);
  for (int i = MAXN - 2; i >= 0; --i) inv[i] = 1ll * inv[i + 1] * (i + 1) % ha;
  int n, m;
  scanf("%d%d", &n, &m);
  int ans = qpow(m, n);
  for (int i = 1; i <= n; ++i) {
    int sm = 0;
    sm = 1ll * qpow(m, n - i + 1) * qpow(2 * m - 1, i - 1) % ha;
    (ans += sm) %= ha;
  }
  printf("%d\n", ans);
  return 0;
}

```