#include <bits/stdc++.h>
const long long modn = 1000000007;
long long inv(long long b) {
  return b == 1 ? 1 : inv(modn % b) * (modn - modn / b) % modn;
}
int main() {
  int a, b;
  int sumbm, suma;
  int ans;
  scanf("%d %d", &a, &b);
  sumbm = (long long)(1 + b - 1) * (b - 1) % modn * inv(2) % modn;
  suma = (long long)(1 + a) * a % modn * inv(2) % modn;
  ans = (long long)sumbm * a % modn;
  ans = ((long long)ans + (long long)b * sumbm % modn * suma % modn) % modn;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in cpp for 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>
const long long modn = 1000000007;
long long inv(long long b) {
  return b == 1 ? 1 : inv(modn % b) * (modn - modn / b) % modn;
}
int main() {
  int a, b;
  int sumbm, suma;
  int ans;
  scanf("%d %d", &a, &b);
  sumbm = (long long)(1 + b - 1) * (b - 1) % modn * inv(2) % modn;
  suma = (long long)(1 + a) * a % modn * inv(2) % modn;
  ans = (long long)sumbm * a % modn;
  ans = ((long long)ans + (long long)b * sumbm % modn * suma % modn) % modn;
  printf("%d\n", ans);
  return 0;
}

```