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 a, b, sumi, sumj;
void add(long long &a, long long b) { a = (a + b) % 1000000007; }
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
int main() {
  ios_base::sync_with_stdio(false);
  cin >> a >> b;
  long long ans = 0;
  for (long long j = 1; j <= b - 1; ++j) add(sumj, j);
  for (long long i = 1; i <= a; ++i) add(sumi, mul(i, sumj));
  ans = (mul(a, sumj) + mul(b, sumi)) % 1000000007;
  cout << ans << endl;
  return 0;
}
