#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, const U &b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, const U &b) {
  if (a < b) a = b;
}
const int mod = 1e9 + 7;
const int NN = 1000100;
long long fac[NN];
int power(int a, int b, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % mod)
    if (b & 1) ans = 1LL * ans * a % mod;
  return ans;
}
int C(int u, int v) {
  long long d =
      fac[u] * power(fac[v], mod - 2) % mod * power(fac[u - v], mod - 2) % mod;
  return d;
}
int main() {
  fac[0] = 1;
  for (int i = 1; i <= 1000000; i++) fac[i] = fac[i - 1] * i % mod;
  int n, m;
  cin >> n >> m;
  long long ans = power(m, n);
  for (int i = 1; i <= n; i++) {
    long long d = power(m, i);
    d *= power(m - 1, n - i);
    d %= mod;
    d *= C(n, i - 1);
    ans += d;
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, const U &b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, const U &b) {
  if (a < b) a = b;
}
const int mod = 1e9 + 7;
const int NN = 1000100;
long long fac[NN];
int power(int a, int b, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % mod)
    if (b & 1) ans = 1LL * ans * a % mod;
  return ans;
}
int C(int u, int v) {
  long long d =
      fac[u] * power(fac[v], mod - 2) % mod * power(fac[u - v], mod - 2) % mod;
  return d;
}
int main() {
  fac[0] = 1;
  for (int i = 1; i <= 1000000; i++) fac[i] = fac[i - 1] * i % mod;
  int n, m;
  cin >> n >> m;
  long long ans = power(m, n);
  for (int i = 1; i <= n; i++) {
    long long d = power(m, i);
    d *= power(m - 1, n - i);
    d %= mod;
    d *= C(n, i - 1);
    ans += d;
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}

```