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() {
  int aa = ({
    int x;
    scanf("%d", &x);
    x;
  }),
      bb = ({
        int x;
        scanf("%d", &x);
        x;
      });
  int ans = 0;
  int mod = 1000000007;
  for (int i = 1; i < bb; i++) {
    int a = i * (bb + 1LL) % mod;
    int n = aa, tt;
    if (n % 2 == 0) {
      tt = (n / 2) * ((2 * a + (n - 1) * (long long)i % mod * bb) % mod) % mod;
    } else {
      tt = n * ((a + (n - 1LL) / 2 * i % mod * bb) % mod) % mod;
    }
    ans = (ans + tt) % mod;
  }
  cout << ans;
}
