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 pow1(long long a, long long b) {
  if (b == 0) return 1;
  long long ret = pow1(a, b / 2);
  ret = (ret * ret) % 1000000007ll;
  if (b & 1) ret = (ret * a) % 1000000007ll;
  return ret;
}
int main() {
  long long a, b;
  scanf("%lld", &a);
  scanf("%lld", &b);
  long long ans = (b * b) % 1000000007ll;
  ans = (ans * (b - 1)) % 1000000007ll;
  ans = (ans * pow1(2, 1000000007ll - 2)) % 1000000007ll;
  ans = (ans * a) % 1000000007ll;
  ans = (ans * (a + 1)) % 1000000007ll;
  ans = (ans * pow1(2, 1000000007ll - 2)) % 1000000007ll;
  long long ans1 = (a * b) % 1000000007ll;
  ans1 = (ans1 * (b - 1)) % 1000000007ll;
  ans1 = (ans1 * pow1(2, 1000000007ll - 2)) % 1000000007ll;
  ans = (ans + ans1) % 1000000007ll;
  printf("%lld\n", ans);
}
