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;
string tostr(int n) {
  stringstream rr;
  rr << n;
  return rr.str();
}
long long qpow(long long n, long long k) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = ans * n;
    n = n * n;
    k >>= 1;
  }
  return ans;
}
const int mod = 1e9 + 7;
const int mxn = 1e5 + 9;
const long double eps = 1e-9;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  long long i, j, n, k, a, b, ans = 0;
  cin >> a >> b;
  ans = (((b * ((a * (a + 1) / 2) % mod)) % mod + a) % mod *
         ((b * (b - 1) / 2) % mod)) %
        mod;
  cout << ans << "\n";
  return 0;
}
