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;
long long a, b, v, ans;
int main() {
  cin >> a >> b;
  for (int i = 1; i <= a; i++) {
    v = b - 1ll;
    if (v % 2 == 0)
      ans =
          (ans + (i * ((((b * (v / 2ll)) % (long long)(1e9 + 7)) * (v + 1ll)) %
                       (long long)(1e9 + 7)))) %
          (long long)(1e9 + 7) % (long long)(1e9 + 7);
    else
      ans =
          (ans + (i * ((((b * ((v + 1ll) / 2ll)) % (long long)(1e9 + 7)) * v) %
                       (long long)(1e9 + 7))) %
                     (long long)(1e9 + 7)) %
          (long long)(1e9 + 7);
    ans = (ans + (((b - 1ll) * b / 2ll)) % (long long)(1e9 + 7)) %
          (long long)(1e9 + 7);
  }
  cout << ans << endl;
  return 0;
}
