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 mod = (int)(1e9 + 7);
long long f(long long a, long x) {
  if (x == 1) {
    return a;
  }
  long long d = f(a, x / 2) % mod;
  if (x % 2 == 0) {
    return (d * d) % mod;
  } else {
    return ((d * d) % mod * a) % mod;
  }
}
int main() {
  long long s, a, b;
  while (cin >> a >> b) {
    s = a;
    for (int k = 1; k <= a; k++) {
      s = s + k * b;
      s = s % mod;
    }
    s = (s * b) % mod;
    s = (s * (b - 1)) % mod;
    s = (s * f(2, mod - 2)) % mod;
    cout << s << endl;
    ;
  }
  return 0;
}
