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;
const int mod = int(1e9) + 7;
long long chen(long long p, long long q) { return p * q % mod; }
int main() {
  int ra, rb;
  scanf("%d%d", &ra, &rb);
  long long a = ra, b = rb;
  long long tmp = a, tmp2 = b;
  if (tmp % 2 == 0)
    tmp = a / 2 * (a + 1);
  else
    tmp = (a + 1) / 2 * a;
  tmp %= mod;
  if (tmp2 % 2 == 0)
    tmp2 = b / 2 * (b - 1);
  else
    tmp2 = (b - 1) / 2 * b;
  tmp2 %= mod;
  long long ans = chen(chen(tmp, b) + a, tmp2);
  printf("%d\n", int(ans));
  return 0;
}
