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 double EPS = 1e-12;
const long long SZ = 2010, SSZ = 21, APB = 26, one = 93,
                INF = 0x3fffffffffffffff, mod = 1000000007;
const int intINF = 0x3f3f3f3f;
double PI = acos(-1);
long long n, m;
int tim;
void init() {
  cin >> n >> m;
  long long res = 0;
  for (int i = 1; i <= m - 1; ++i) {
    res += n * (n + 1) / 2 % mod * i % mod * m % mod + n * i % mod;
    res %= mod;
  }
  res = (res % mod + mod) % mod;
  cout << res << endl;
}
void work() {}
int main() {
  std::ios::sync_with_stdio(0);
  long long casenum;
  {
    init();
    work();
  }
  return 0;
}
