#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int t, i, j, k, l, m, n, sum, a, b;
  scanf("%lld%lld", &a, &b);
  sum = 0;
  for (i = 1; i <= a; i++) {
    j = b * i + 1;
    if (j % b != 0) sum = (sum + j) % 1000000007;
  }
  b = (b * (b - 1) / 2) % 1000000007;
  sum = (sum * b) % 1000000007;
  printf("%lld", sum % 1000000007);
  return 0;
}

### Prompt

Please formulate 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;
int main() {
  long long int t, i, j, k, l, m, n, sum, a, b;
  scanf("%lld%lld", &a, &b);
  sum = 0;
  for (i = 1; i <= a; i++) {
    j = b * i + 1;
    if (j % b != 0) sum = (sum + j) % 1000000007;
  }
  b = (b * (b - 1) / 2) % 1000000007;
  sum = (sum * b) % 1000000007;
  printf("%lld", sum % 1000000007);
  return 0;
}

```