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;
void smain();
int main() {
  ios_base::sync_with_stdio(0);
  smain();
  return 0;
}
long long a, b;
long long solveb() {
  long long res = 0;
  set<long long> vals;
  for (long long p = 1; p < b; ++p) {
    for (long long k = 1; k <= a; ++k) vals.insert((b * k + 1) * p);
  }
  for (auto it = vals.begin(); it != vals.end(); ++it) res += *it;
  return res;
}
void smain() {
  cin >> a >> b;
  long long res = 0;
  long long mul = (1 + a) * a / 2;
  mul %= 1000000007;
  mul = mul * b % 1000000007;
  mul += a;
  if (mul >= 1000000007) mul -= 1000000007;
  for (long long p = 1; p < b; ++p) {
    res = (p * mul + res) % 1000000007;
  }
  cout << res << '\n';
}
