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 = 1000000007;
const int MAXN = 1000006;
long long a, b;
long long suma(long long e, long long l) {
  long long m1 = e, m2 = e + 1;
  if (m1 % 2 == 0)
    m1 /= 2;
  else
    m2 /= 2;
  m1 %= mod;
  m2 %= mod;
  e %= mod;
  return (m1 * m2 % mod * b % mod * l % mod + l * (e + 1)) % mod;
}
int main() {
  scanf("%lld%lld", &a, &b);
  long long ret = 0;
  for (int l = 1; l < b; ++l) {
    ret += suma(a, l);
    ret += mod - l;
  }
  printf("%lld\n", ret % mod);
  return 0;
}
