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-11;
const long long MOD = 1000000007;
const int N = 1000000 + 7;
void fast() { std::ios_base::sync_with_stdio(0); }
long long f(long long n) {
  long long x = n;
  long long y = n + 1;
  if (x % 2 == 0)
    x /= 2;
  else
    y /= 2;
  x %= MOD;
  y %= MOD;
  return (x * y) % MOD;
}
int main() {
  long long a, b;
  cin >> a >> b;
  long long Asum = f(a);
  long long Bsum = f(b - 1);
  Asum = (Asum * b) % MOD;
  Asum = (Asum + a) % MOD;
  Asum = (Asum * Bsum) % MOD;
  cout << Asum;
  return 0;
}
