#include <bits/stdc++.h>
using namespace std;
const int N = (int)1e5 + 7;
const int inf = (int)1e9 + 7;
int mod = 1e9 + 7;
long long mult(long long a, long long b) { return a * b % mod; }
long long get(long long x, int id) { return x * (x + id) / 2 % mod; }
int main() {
  int a, b;
  scanf("%d %d", &a, &b);
  printf("%lld", mult((mult(get(a, 1), b) + a), get(b, -1)));
}

### 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 int N = (int)1e5 + 7;
const int inf = (int)1e9 + 7;
int mod = 1e9 + 7;
long long mult(long long a, long long b) { return a * b % mod; }
long long get(long long x, int id) { return x * (x + id) / 2 % mod; }
int main() {
  int a, b;
  scanf("%d %d", &a, &b);
  printf("%lld", mult((mult(get(a, 1), b) + a), get(b, -1)));
}

```