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}.
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long a, b;
  scanf("%I64d%I64d", &a, &b);
  long long res1 = a * b + 1 + b + 1;
  long long res2 = a;
  if (res1 % 2)
    res2 /= 2;
  else
    res1 /= 2;
  long long res3 = (b - 1);
  long long res4 = b;
  if (res3 % 2)
    res4 /= 2;
  else
    res3 /= 2;
  res1 %= 1000000007;
  res2 %= 1000000007;
  long long ans =
      res1 * res2 % 1000000007 * res3 % 1000000007 * res4 % 1000000007;
  printf("%I64d", ans % 1000000007);
  return 0;
}
