#include <bits/stdc++.h>
using namespace std;
long long a, b;
int main() {
  cin >> a >> b;
  cout << ((((((a * (a + 1) / 2) % 1000000007) * b) % 1000000007 + a) %
            1000000007) *
           ((b * (b - 1) / 2) % 1000000007)) %
              1000000007;
  return 0;
}

### Prompt

Generate 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;
long long a, b;
int main() {
  cin >> a >> b;
  cout << ((((((a * (a + 1) / 2) % 1000000007) * b) % 1000000007 + a) %
            1000000007) *
           ((b * (b - 1) / 2) % 1000000007)) %
              1000000007;
  return 0;
}

```