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;
const long long M = 1e9 + 7;
long long a, b;
long long ans = 0;
long long a1, b1;
void init() { cin >> a >> b; }
void work() {
  a1 = (a % 2 == 0) ? ((a / 2) * (a + 1)) : (a * ((a + 1) / 2));
  a1 %= M;
  b1 = (b % 2 == 0) ? ((b / 2) * (b - 1)) : (b * ((b - 1) / 2));
  b1 %= M;
  ans = (((a1 * b1) % M) * b) % M;
  ans = (ans + a * b1) % M;
  cout << ans << endl;
}
int main() {
  init();
  work();
  return 0;
}
