#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int a;
  long long int b;
  scanf("%lld", &a);
  scanf("%lld", &b);
  long long int val1 = (b * (b - 1)) / 2;
  val1 = val1 % 1000000007;
  long long int val2 = (a * (a + 1)) / 2;
  val2 = val2 % 1000000007;
  val2 = val2 * b;
  val2 = val2 % 1000000007;
  val2 = (val2 + a) % 1000000007;
  long long int ans = val1 * val2;
  ans = ans % 1000000007;
  cout << ans << endl;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 int a;
  long long int b;
  scanf("%lld", &a);
  scanf("%lld", &b);
  long long int val1 = (b * (b - 1)) / 2;
  val1 = val1 % 1000000007;
  long long int val2 = (a * (a + 1)) / 2;
  val2 = val2 % 1000000007;
  val2 = val2 * b;
  val2 = val2 % 1000000007;
  val2 = (val2 + a) % 1000000007;
  long long int ans = val1 * val2;
  ans = ans % 1000000007;
  cout << ans << endl;
}

```