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>
char ch;
inline void read(int &x) {
  x = 0;
  ch = getchar();
  while (ch <= 32) ch = getchar();
  int f = 1;
  if (ch == '-') f = -1, ch = getchar();
  while (ch > 32) x = x * 10 + ch - 48, ch = getchar();
  x *= f;
};
int main() {
  int a, b;
  read(a);
  read(b);
  int ans =
      (1LL * b * (b - 1) / 2LL % 1000000007) *
      (a + 1LL * b * (1LL * a * (a + 1) / 2LL % 1000000007) % 1000000007) %
      1000000007;
  printf("%d\n", ans);
}
