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;
void solve() {
  int a, b;
  cin >> a >> b;
  int mod = ((a - 1) % b + b) % b;
  long long int sum = 0;
  for (int i = a; i <= b; i++) {
    int num = i / b;
    int den = mod + 1;
    if (den >= b) den -= b;
    mod = den;
    if (den == 0) continue;
    if (num % den != 0) continue;
    if (num / den >= 1 && num / den <= a) {
      sum += i;
    }
  }
  printf("%lld\n", sum);
}
void solve1() {
  long long int mod = 1000000007;
  long long int a, b;
  cin >> a >> b;
  long long int sum = 0;
  for (long long int t = 1; t < b; t++) {
    sum =
        (sum + ((b * t) % mod * (((a * (a + 1L)) / 2L) % mod)) % mod + t * a) %
        mod;
    assert(sum >= 0);
  }
  printf("%lld\n", sum);
}
int main() {
  int T = 1;
  for (int i = 1; i <= T; i++) {
    solve1();
  }
  return 0;
}
