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;
  while (scanf("%lld%lld", &a, &b) != -1) {
    long long t, t1;
    long long sum = 0;
    long long k1, k2;
    for (t1 = 1; t1 < b; t1++) {
      if (a % 2 == 0) {
        k1 = a / 2;
        k2 = (a * t1 + t1);
      } else {
        k1 = (a * t1 + t1) / 2;
        k2 = a;
      }
      long long p = ((((((k1 % 1000000007) * (k2 % 1000000007)) % 1000000007) %
                       1000000007) *
                      (b % 1000000007)) %
                         1000000007 +
                     (((a % 1000000007) * (t1 % 1000000007)) % 1000000007)) %
                    1000000007;
      sum = sum % 1000000007;
      sum = (p + sum) % 1000000007;
    }
    cout << sum << endl;
  }
}
