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 MOD = 1000 * 1000 * 1000 + 7;
int add(int a, int b) { return ((a + b) % MOD + MOD) % MOD; }
int mul(int a, int b) { return ((long long)a * b) % MOD; }
int main() {
  int a, b;
  cin >> a >> b;
  int ans = 0;
  for (int r = 1; r < b; r++) {
    if (a & 1) {
      ans = add(ans, mul((a + 1) / 2, mul(a, mul(r, b))));
    } else {
      ans = add(ans, mul(a / 2, mul(a + 1, mul(r, b))));
    }
    ans = add(ans, mul(a, r));
  }
  cout << ans << endl;
  return 0;
}
