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, mod, sumz, sumy, sum, last, first, kol, z;
long long run(long long x, long long step) {
  long long res = 1;
  while (step) {
    if (step & 1) res = (res * x) % mod;
    x = x * x % mod;
    step >>= 1;
  }
  return res % mod;
}
int main() {
  cin >> a >> b;
  mod = 1000 * 1000 * 1000 + 7;
  sumz = (a * b) % mod;
  sumz = (sumz * (b - 1)) % mod;
  sumz = (sumz * run(2, mod - 2)) % mod;
  sumy = (a * (a + 1)) % mod;
  sumy = (sumy * b) % mod;
  sumy = (sumy * (b - 1)) % mod;
  sumy = (sumy * run(4, mod - 2)) % mod;
  sum = ((b * sumy) % mod + sumz) % mod;
  cout << sum << endl;
  return 0;
}
