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;
inline int nextInt() {
  char ch;
  int x = 0;
  bool flag = false;
  do ch = (char)getchar(), flag = (ch == '-') ? true : flag;
  while (ch < '0' || ch > '9');
  do x = x * 10 + ch - '0';
  while (ch = (char)getchar(), ch <= '9' && ch >= '0');
  return x * (flag ? -1 : 1);
}
int n, m;
int main() {
  int i, j, k;
  int x, y, z;
  long long a, b;
  scanf("%lld%lld", &a, &b);
  long long ans;
  ans = b * (b - 1) / 2 % 1000000007 * a + a * (a + 1) / 2 % 1000000007 *
                                               (b * (b - 1) / 2 % 1000000007) %
                                               1000000007 * b % 1000000007;
  ans %= 1000000007;
  printf("%lld", ans);
  return 0;
}
