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 GCD(int a, int b) {
  if (b == 0) return a;
  return (a % b == 0 ? b : GCD(b, a % b));
}
long long int POW(long long int base, long long int exp) {
  long long int val;
  val = 1;
  while (exp > 0) {
    if (exp % 2 == 1) {
      val = (val * base) % 1000000007;
    }
    base = (base * base) % 1000000007;
    exp = exp / 2;
  }
  return val;
}
int main() {
  int a, b;
  long long mul, val;
  scanf("%d%d", &a, &b);
  mul = (b * 1LL * (b - 1) / 2) % 1000000007;
  val = (a * 1LL * (a + 1) / 2) % 1000000007;
  val = (val * b) % 1000000007;
  val = (val + a) % 1000000007;
  val = (val * mul) % 1000000007;
  cout << val << endl;
  return 0;
}
