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 long long mod = (long long)1e9 + 7;
long long mul(long long a, long long b) {
  a %= mod, b %= mod;
  return a * b % mod;
}
long long add(long long a, long long b) {
  a %= mod, b %= mod;
  a += b, a %= mod;
  return a;
}
int main(int argc, char const *argv[]) {
  long long a, b;
  cin >> a >> b;
  long long res = 0;
  for (int(q) = (1); (q) < (int)(b); ++(q)) {
    res = add(res, mul(mul(q, b), a * (a + 1) / 2));
    res = add(res, a * q);
  }
  cout << res << endl;
  return 0;
}
