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 P = 1000000007;
int a, b, mo, ans, _2;
int Pow(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = 1ll * a * a % P)
    if (b & 1) res = 1ll * res * a % P;
  return res;
}
int calc(int a0, int d, int n) {
  int an = a0 + 1ll * d * (n - 1) % P;
  an %= P;
  int res = (a0 + an) % P * 1ll * n % P * 1ll * _2 % P;
  return res;
}
int main() {
  scanf("%d%d", &a, &b);
  _2 = Pow(2, P - 2);
  for (mo = 1; mo < b; mo++) {
    ans += calc((mo + 1ll * b * mo) % P, (1ll * b * mo) % P, a);
    if (ans >= P) ans -= P;
  }
  printf("%d\n", ans);
}
