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;
template <typename T>
using V = vector<T>;
const long long mod = 1000000007;
long long pow_mod(long long base, long long power) {
  long long res = 1;
  while (power) {
    if (power & 1) res = (res * base) % mod;
    base = (base * base) % mod;
    power >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long a, b;
  cin >> a >> b;
  long long res = 0;
  long long fs = 0, sc = 0;
  sc = (((b * (b - 1)) % mod) * pow_mod(2, mod - 2)) % mod;
  fs = (sc * b) % mod;
  for (__typeof(a + 1) k = (1) - ((1) > (a + 1));
       k != (a + 1) - ((1) > (a + 1)); k += ((1) > (a + 1) ? -1 : 1)) {
    res = (res + k * fs + sc) % mod;
  }
  cout << res << '\n';
  return 0;
}
