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 int mod = (int)1e9 + 7;
const int d = mod / 2 + 1;
int add(int a, int b) { return (a + b) % mod; }
int mul(int a, int b) { return (long long int)a * b % mod; }
int main() {
  long long int a, b;
  scanf("%lld%lld", &a, &b);
  long long int ans = 0;
  for (long long int m = 1; m < b; m++) {
    long long int sum_t = mul(mul(mul(m, a + 1), a), d);
    long long int cur_add = mul(b, sum_t) + mul(m, a);
    cur_add %= mod;
    ans = add(ans, cur_add);
  }
  printf("%lld\n", ans);
  return 0;
}
