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;
using ll = long long;
using pii = pair<int, int>;
using tpl = tuple<int, int, int>;
using pcc = pair<char, char>;
const ll inf = 2e18;
const ll mod = 1e9 + 7;
int dr[] = {-1, 0, 1, 0};
int dc[] = {0, 1, 0, -1};
int kr[] = {-1, 1, -2, -2, -1, 1, 2, 2};
int kc[] = {-2, -2, -1, 1, 2, 2, -1, 1};
int dgr[] = {-1, -1, 1, 1};
int dgc[] = {1, -1, -1, 1};
int main() {
  ll a, b;
  cin >> a >> b;
  ll ans = 0, sum = 0;
  sum = b * (b - 1) / 2;
  sum %= mod;
  for (ll k = 1; k <= a; k++) {
    ans +=
        (((((k % mod) * (b % mod)) % mod + 1) % mod) % mod * sum % mod) % mod;
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}
