#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
const int mod = 1e9 + 7;
int n, m;
long long ans;
long long fac[maxn], inv[maxn], f[maxn];
long long power(long long b, int k) {
  long long res = 1;
  for (; k; k >>= 1, b = b * b % mod)
    if (k & 1) res = res * b % mod;
  return res;
}
long long C(int n, int m) { return fac[n] * inv[m] % mod * inv[n - m] % mod; }
int main() {
  long long s = 1;
  scanf("%d%d", &n, &m);
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
  inv[n] = power(fac[n], mod - 2);
  for (int i = n - 1; i >= 1; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
  f[0] = 1;
  for (int i = 1; i <= n; i++) f[i] = f[i - 1] * m % mod;
  ans = f[n];
  for (int i = n; i >= 1; i--) f[i] = f[i] * s % mod, s = s * (m - 1) % mod;
  for (int i = 1; i <= n; i++) ans = (ans + f[i] * C(n, i - 1)) % mod;
  printf("%I64d\n", ans);
  return 0;
}

### 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 int maxn = 1e6 + 5;
const int mod = 1e9 + 7;
int n, m;
long long ans;
long long fac[maxn], inv[maxn], f[maxn];
long long power(long long b, int k) {
  long long res = 1;
  for (; k; k >>= 1, b = b * b % mod)
    if (k & 1) res = res * b % mod;
  return res;
}
long long C(int n, int m) { return fac[n] * inv[m] % mod * inv[n - m] % mod; }
int main() {
  long long s = 1;
  scanf("%d%d", &n, &m);
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
  inv[n] = power(fac[n], mod - 2);
  for (int i = n - 1; i >= 1; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
  f[0] = 1;
  for (int i = 1; i <= n; i++) f[i] = f[i - 1] * m % mod;
  ans = f[n];
  for (int i = n; i >= 1; i--) f[i] = f[i] * s % mod, s = s * (m - 1) % mod;
  for (int i = 1; i <= n; i++) ans = (ans + f[i] * C(n, i - 1)) % mod;
  printf("%I64d\n", ans);
  return 0;
}

```