#include <bits/stdc++.h>
using namespace std;
using namespace std;
bool debug = 0;
void run();
int main() {
  int st = clock();
  run();
  int en = clock();
  freopen("CON", "w", stdout);
  return 0;
}
int a, b;
long long s(long long n) { return (((n) * (n + 1)) / 2) % 1000000007; }
void run() {
  scanf("%d %d", &a, &b);
  long long sum = ((((b * s(a)) % 1000000007) * s(b - 1)) % 1000000007 +
                   (s(b - 1) * a) % 1000000007) %
                  1000000007;
  printf("%I64d ", sum);
  printf("\n");
}

### Prompt

Please create a solution in cpp to the following problem:

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;
using namespace std;
bool debug = 0;
void run();
int main() {
  int st = clock();
  run();
  int en = clock();
  freopen("CON", "w", stdout);
  return 0;
}
int a, b;
long long s(long long n) { return (((n) * (n + 1)) / 2) % 1000000007; }
void run() {
  scanf("%d %d", &a, &b);
  long long sum = ((((b * s(a)) % 1000000007) * s(b - 1)) % 1000000007 +
                   (s(b - 1) * a) % 1000000007) %
                  1000000007;
  printf("%I64d ", sum);
  printf("\n");
}

```