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 power(long long a, long long b) {
  long long x = a;
  long long y = b;
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    x = x * x;
    y >>= 1;
  }
  return res;
}
long long invmod(long long a) { return power(a, 1000000007 - 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long a, b;
  cin >> a >> b;
  long long ans, temp;
  ans = (a * (a + 1)) / 2;
  ans %= 1000000007;
  temp = (b * (b - 1)) / 2;
  temp %= 1000000007;
  temp *= b;
  temp %= 1000000007;
  ans = (ans * temp) % 1000000007;
  temp = (b * (b - 1)) / 2;
  temp %= 1000000007;
  temp *= a;
  temp %= 1000000007;
  ans = (ans + temp) % 1000000007;
  cout << ans << endl;
  return 0;
}
