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;
template <typename TYPE>
inline TYPE sqr(const TYPE &a) {
  return (a) * (a);
}
const int INF = 1000 * 1000 * 1000;
const long double EPS = 1e-9;
const long double PI = 2 * acos(0.0);
const int N = 100;
const int MODULO = INF + 7;
void add(long long &to, long long v) { to = (to + v) % MODULO; }
void mult(long long &to, long long v) { to = (to * v) % MODULO; }
int main() {
  long long a, b;
  cin >> a >> b;
  long long ans = (a * (a + 1)) / 2;
  mult(ans, 1);
  long long bb = (b * (b - 1)) / 2;
  mult(bb, 1);
  long long x = bb;
  mult(x, b);
  mult(ans, x);
  x = a;
  mult(x, bb);
  add(ans, x);
  cout << ans << endl;
  return 0;
}
