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 mod = 1e9 + 7;
long long inv2 = (mod + 1) / 2;
long long sm2(long long x) {
  if (x >= mod) x %= mod;
  return (((x * (x + 1)) % mod) * inv2) % mod;
}
int main() {
  long long a, b;
  cin >> a >> b;
  long long sm1 = 0, sm = 0;
  for (long long l = 1; l < b; l++) {
    long long mx = a;
    long long mn = 1;
    sm1 += (l * (sm2(mx) - sm2(mn - 1))) % mod;
    sm = (sm + ((mx - mn + 1) % mod) * l) % mod;
  }
  sm1 = (sm1 % mod + mod) % mod;
  cout << (b * sm1 + sm) % mod;
}
