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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int inf = 1e9 + 7;
const long long linf = 1e18;
const double pi = acos(-1.0);
template <class T>
T sqr(T x) {
  return x * x;
}
const double eps = 1e-6;
long long fastpow(int m, int n) {
  int temp = n;
  long long ans = 1;
  long long t = m;
  while (temp > 0) {
    if (temp & 1) ans = (ans * t) % inf;
    t = (t * t) % inf;
    temp = temp >> 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  long long a, b;
  cin >> a >> b;
  long long ni = fastpow(2, inf - 2);
  long long ans = b * (b - 1) % inf * ni % inf;
  long long ans1 = a * (a + 1) % inf * ni % inf * b % inf;
  ans1 = (ans1 + a) % inf;
  ans = (ans * ans1) % inf;
  cout << ans << endl;
  return 0;
}
