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() {
  ios_base::sync_with_stdio(false);
  uint32_t a, b;
  cin >> a >> b;
  uint32_t res = 0;
  uint64_t sumB = (uint64_t(b) * (b - 1) / 2) % 1000000007;
  for (uint64_t i = 1; i <= a; ++i) {
    uint64_t val = ((i * b + 1) % 1000000007) * sumB;
    res = (uint32_t)((res + val) % 1000000007);
  }
  cout << res << endl;
  return 0;
}
