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;
long long a, b;
long long bigmod(long long base, long long pow) {
  if (pow == 0) return 1;
  long long p = bigmod(base, pow / 2);
  p %= 1000000007;
  p *= p;
  p %= 1000000007;
  if (pow % 2) p *= base;
  p %= 1000000007;
  return p;
}
long long fun() {
  long long my = b;
  my %= 1000000007;
  my *= (b - 1);
  my %= 1000000007;
  long long lol = a * b;
  lol %= 1000000007;
  lol *= (a + 1);
  lol %= 1000000007;
  lol += 2 * a;
  lol %= 1000000007;
  my *= lol;
  my %= 1000000007;
  my = (my % 1000000007 * bigmod(4, 1000000007 - 2));
  my %= 1000000007;
  return my;
}
int main() {
  while (scanf("%lld %lld", &a, &b) != EOF) {
    long long ans = fun();
    printf("%lld", ans);
    printf("\n");
  }
  return 0;
}
