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 int exp(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long e;
  e = exp(a, b / 2);
  e = ((e * e) % 1000000007);
  if (b % 2 == 0)
    return e;
  else
    return ((e * a) % 1000000007);
}
int main() {
  long long int a, b;
  cin >> a >> b;
  long long int sum = 0;
  sum = (((((b - 1) * b) % 1000000007) *
          ((a * (((a * b + b + 2) % 1000000007))) % 1000000007)) %
         1000000007);
  sum = ((sum * exp((long long int)4, (long long int)1000000005)) % 1000000007);
  cout << sum << endl;
  return 0;
}
