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 MAXN = 100000 + 2;
const int MOD = (int)(1e9) + 7;
long long m;
long long power(int x, int k) {
  if (k == 1) return (x);
  long long t = power(x, k / 2);
  t = (t * t) % MOD;
  if (k % 2 == 1) t = (t * x) % MOD;
  return (t);
}
long long sum(long long n) {
  n = n % MOD;
  long long res = (n * (n + 1)) % MOD;
  res = (res * m) % MOD;
  return res;
}
int main() {
  long long a, b;
  cin >> a >> b;
  m = power(2, MOD - 2);
  long long v = sum(a);
  long long res = 0;
  for (int i = 1; i < b; i++) {
    long long l = 1, r = (long long)(a);
    int y = (long long)(b)*i % MOD;
    res = ((res + (a * i) % MOD + v * y) % MOD) % MOD;
  }
  res = (res + MOD) % MOD;
  cout << res << endl;
}
