#include <bits/stdc++.h>
using namespace std;
int a, b;
long long result;
int main() {
  while (cin >> a >> b) {
    result = 0;
    for (long long i = 1; i <= a; i++) {
      long long temp;
      temp = ((i * b) + 1) % 1000000007;
      if (b % 2) {
        temp *= (b - 1) / 2;
        temp %= 1000000007;
        temp *= b;
        temp %= 1000000007;
      } else {
        temp *= b / 2;
        temp %= 1000000007;
        temp *= (b - 1);
        temp %= 1000000007;
      }
      result += temp;
      result %= 1000000007;
    }
    cout << result << endl;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int a, b;
long long result;
int main() {
  while (cin >> a >> b) {
    result = 0;
    for (long long i = 1; i <= a; i++) {
      long long temp;
      temp = ((i * b) + 1) % 1000000007;
      if (b % 2) {
        temp *= (b - 1) / 2;
        temp %= 1000000007;
        temp *= b;
        temp %= 1000000007;
      } else {
        temp *= b / 2;
        temp %= 1000000007;
        temp *= (b - 1);
        temp %= 1000000007;
      }
      result += temp;
      result %= 1000000007;
    }
    cout << result << endl;
  }
  return 0;
}

```