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 MOD = 1000000007;
int a, b;
long long res;
long long get_sum(long long x) {
  x %= MOD;
  return (x * (x + 1LL) / 2LL) % MOD;
}
int main() {
  scanf("%d %d", &a, &b);
  for (int i = 1; i <= a; ++i) {
    long long tmp = b * (long long)i + 1LL;
    tmp %= MOD;
    res += get_sum(b - 1) * tmp;
    res %= MOD;
  }
  printf("%I64d\n", res % MOD);
  return 0;
}
