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

### Prompt

Create a solution in CPP for 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;
const long long maxn = 2e6 + 10;
const long long mod = 1e9 + 7;
long long fact[maxn];
long long pow(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    long long y = pow(a, b / 2);
    return (y * y) % mod;
  } else {
    return (pow(a, b - 1) * a) % mod;
  }
}
long long C(long long n, long long k) {
  return fact[n] * pow((fact[n - k] * fact[k] % mod), mod - 2) % mod;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n, m;
  cin >> n >> m;
  fact[0] = 1;
  for (long long i = 1; i < maxn; ++i) {
    fact[i] = fact[i - 1] * i % mod;
  }
  long long ans = 0;
  for (long long sum = 1; sum <= n; ++sum) {
    ans += pow(m, sum) * pow(m - 1, n - sum) % mod * C(n, sum - 1) % mod;
    ans %= mod;
  }
  cout << (ans + pow(m, n)) % mod;
}

```