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() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int t;
  t = 1;
  while (t--) {
    long long a, b, ans = 0;
    cin >> a >> b;
    for (long long i = 1; i < (a + 1); ++i) {
      long long ch1 = ((b * (b - 1)) / 2) % 1000000007;
      long long ch = (ch1 * ((i * b + 1) % 1000000007)) % 1000000007;
      ans = (ans % 1000000007 + ch % 1000000007) % 1000000007;
    }
    cout << ans << endl;
  }
}
