#include <bits/stdc++.h>
using namespace std;
int main() {
  long long a, b, wyn = 0, m = 1000000007, c, d;
  cin >> a >> b;
  d = (1 + a) * a / 2;
  d %= m;
  for (int i = 1; i < b; i++) {
    c = i * b;
    c %= m;
    c *= d;
    c %= m;
    c += i * a;
    c %= m;
    wyn += c;
    wyn %= m;
  }
  cout << wyn;
}

### Prompt

Please create a solution in Cpp 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;
int main() {
  long long a, b, wyn = 0, m = 1000000007, c, d;
  cin >> a >> b;
  d = (1 + a) * a / 2;
  d %= m;
  for (int i = 1; i < b; i++) {
    c = i * b;
    c %= m;
    c *= d;
    c %= m;
    c += i * a;
    c %= m;
    wyn += c;
    wyn %= m;
  }
  cout << wyn;
}

```