#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long N = 1e6 + 5;
long long a, b;
long long MOD(long long x) { return x % mod; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  cin >> a >> b;
  long long cnta = MOD(a * (a + 1) / 2), cntb = MOD(b * (b - 1) / 2);
  cout << MOD(MOD(MOD(cnta * b + a) * cntb));
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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}.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long N = 1e6 + 5;
long long a, b;
long long MOD(long long x) { return x % mod; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  cin >> a >> b;
  long long cnta = MOD(a * (a + 1) / 2), cntb = MOD(b * (b - 1) / 2);
  cout << MOD(MOD(MOD(cnta * b + a) * cntb));
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

```