#include <bits/stdc++.h>
using namespace std;
long long n, m, k;
long long read() {
  long long s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}
inline void print(long long *f, long long len) {
  for (long long i = 0; i < len; i++) printf("%lld ", f[i]);
  puts("");
}
const long long mod = 1e9 + 7;
long long qpow(long long a, long long b = mod - 2) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
signed main() {
  cin >> n >> m;
  long long ans = qpow(m, n), inv = qpow(m), invv = ans;
  long long nw = 1;
  for (long long i = 1, iend = n; i <= iend; ++i)
    ans += invv * nw % mod, ans %= mod, invv *= inv, invv %= mod,
        nw *= (m - 1 + m), nw %= mod;
  cout << ans;
}

### 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;
long long n, m, k;
long long read() {
  long long s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}
inline void print(long long *f, long long len) {
  for (long long i = 0; i < len; i++) printf("%lld ", f[i]);
  puts("");
}
const long long mod = 1e9 + 7;
long long qpow(long long a, long long b = mod - 2) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
signed main() {
  cin >> n >> m;
  long long ans = qpow(m, n), inv = qpow(m), invv = ans;
  long long nw = 1;
  for (long long i = 1, iend = n; i <= iend; ++i)
    ans += invv * nw % mod, ans %= mod, invv *= inv, invv %= mod,
        nw *= (m - 1 + m), nw %= mod;
  cout << ans;
}

```