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 double pi = acos(-1.0);
const double eps = 1e-10;
const long long MOD = 1e9 + 7;
long long farm(int a, int b) {
  long long i, j, k;
  long long k1a = (1 + a);
  k1a *= a;
  k1a /= 2;
  k1a %= MOD;
  long long j1b = (1 + b - 1);
  j1b *= b - 1;
  j1b /= 2;
  j1b %= MOD;
  j1b *= k1a;
  j1b %= MOD;
  long long ans = 1;
  long long ta = (1LL + (long long)a) * a / 2;
  ta %= MOD;
  ta = ta * b + a;
  ta %= MOD;
  long long tb = (long long)b * (b - 1) / 2;
  tb %= MOD;
  ans = ta * tb;
  ans %= MOD;
  return ans % MOD;
}
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  printf("%I64d\n", farm(a, b));
  return 0;
}
