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 <class T>
void DBGpair(T a, string s = "") {
  return;
}
template <class T>
void DBGvec(vector<T> &a) {}
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
char dir[] = {'R', 'L', 'D', 'U'};
const int MOD = 1e9 + 7;
const long long INF = 1e9;
void setIO(string filename) {
  if (fopen((filename + ".in").c_str(), "r")) {
    freopen((filename + ".in").c_str(), "r", stdin);
    freopen((filename + ".out").c_str(), "w", stdout);
  }
}
void solve() {
  long long a, b;
  cin >> a >> b;
  long long B = (b * (b - 1) / 2) % MOD;
  long long A1 = (a * (a + 1) / 2) % MOD;
  long long A = (A1 * b + a) % MOD;
  long long answer = (A * B) % MOD;
  cout << answer << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) solve();
  return 0;
}
