#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
long long Pow(long long b, long long p = mod - 2) {
  if (p == 0) return 1;
  long long z = Pow(b, p / 2);
  z = (z * z) % mod;
  if (p % 2 == 1) z = (z * b) % mod;
  return z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long a, b;
  cin >> a >> b;
  long long r, q, x1, x2;
  x1 = (b * (b - 1)) % mod;
  x1 *= Pow(2);
  x1 %= mod;
  x2 = (x1 * (2 * b - 1)) % mod;
  x2 *= Pow(3);
  x2 %= mod;
  r = (x1 * a) % mod;
  q = (a * (a + 1)) % mod;
  q = (q * x1) % mod;
  q *= Pow(2);
  q %= mod;
  long long ans;
  ans = (q * b) % mod;
  ans = (ans + r) % mod;
  cout << ans << "\n";
}

### Prompt

Your task is to create a CPP solution to the following problem:

Dreamoon loves summing up something for no reason. One day he obtains two integers a and b occasionally. He wants to calculate the sum of all nice integers. Positive integer x is called nice if <image> and <image>, where k is some integer number in range [1, a].

By <image> we denote the quotient of integer division of x and y. By <image> we denote the remainder of integer division of x and y. You can read more about these operations here: http://goo.gl/AcsXhT.

The answer may be large, so please print its remainder modulo 1 000 000 007 (109 + 7). Can you compute it faster than Dreamoon?

Input

The single line of the input contains two integers a, b (1 ≤ a, b ≤ 107).

Output

Print a single integer representing the answer modulo 1 000 000 007 (109 + 7).

Examples

Input

1 1


Output

0


Input

2 2


Output

8

Note

For the first sample, there are no nice integers because <image> is always zero.

For the second sample, the set of nice integers is {3, 5}.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
long long Pow(long long b, long long p = mod - 2) {
  if (p == 0) return 1;
  long long z = Pow(b, p / 2);
  z = (z * z) % mod;
  if (p % 2 == 1) z = (z * b) % mod;
  return z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long a, b;
  cin >> a >> b;
  long long r, q, x1, x2;
  x1 = (b * (b - 1)) % mod;
  x1 *= Pow(2);
  x1 %= mod;
  x2 = (x1 * (2 * b - 1)) % mod;
  x2 *= Pow(3);
  x2 %= mod;
  r = (x1 * a) % mod;
  q = (a * (a + 1)) % mod;
  q = (q * x1) % mod;
  q *= Pow(2);
  q %= mod;
  long long ans;
  ans = (q * b) % mod;
  ans = (ans + r) % mod;
  cout << ans << "\n";
}

```