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 = 1000000007;
int a, b;
int cheng(int x, int y) {
  long long res = x;
  res *= y;
  res %= MOD;
  return res;
}
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  int la, lb;
  if (a & 1)
    la = cheng(a, a / 2 + 1);
  else
    la = cheng(a / 2, a + 1);
  if (b & 1)
    lb = cheng(b, b / 2);
  else
    lb = cheng(b / 2, b - 1);
  int res = cheng(lb, a);
  res += cheng(cheng(la, lb), b);
  res %= MOD;
  printf("%d\n", res);
  return 0;
}
