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 = 1000000007;
int m, n, p;
char arr[20], arr2[20];
int func(int i, int pos) {
  if (i == n) {
    if (pos == p)
      return 1;
    else
      return 0;
  }
  int ans = 0;
  if (arr2[i] == '+') {
    ans = func(i + 1, pos + 1);
  } else if (arr2[i] == '-')
    ans = func(i + 1, pos - 1);
  else
    ans = func(i + 1, pos + 1) + func(i + 1, pos - 1);
  return ans;
}
int main() {
  long long i, t, k, l, a, b, q;
  long long ans = 0;
  double temp;
  cin >> a >> b;
  ans = ((b * (b - 1)) / 2) % mod;
  long long ans2 = ((a * (a + 1)) / 2) % mod;
  ans2 = (ans2 * b) % mod;
  ans2 = (ans2 + a) % mod;
  ans = (ans * ans2) % mod;
  cout << ans;
  return 0;
}
