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 t, d, ct, q, ans, l, m, r, n, mx, mn, i, j, k, x, y, z, s;
char ch;
string ss = "", str = "";
map<long long, long long> mp;
set<long long> st;
long long a[200003], b[200003];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  s = k = d = l = mx = m = mn = r = q = i = j = z = x = y = ct = ans = 0;
  cin >> x >> y;
  i = (y * (y - 1) / 2) % 1000000007;
  j = (x + ((x * (x + 1) / 2) % 1000000007 * y) % 1000000007);
  cout << (i * j) % 1000000007;
  return 0;
}
