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;
template <class t, class u>
void chmax(t& first, u second) {
  if (first < second) first = second;
}
template <class t, class u>
void chmin(t& first, u second) {
  if (second < first) first = second;
}
const long long int MOD = 1e9 + 7;
inline long long int sum(long long int k) { return (k * (k + 1) / 2) % MOD; }
int main() {
  long long int A, B;
  scanf("%lld", &A);
  scanf("%lld", &B);
  long long int ans = 0LL;
  for (int i = int(1); i < int(B); i++) {
    long long int k = A * i;
    long long int num = A;
    long long int sm1 = (sum(A) * i) % MOD;
    ans = (ans + (B * sm1) % MOD) % MOD;
    ans = (ans + (i * num) % MOD) % MOD;
  }
  printf("%lld ", ans);
  return 0;
}
