#include <bits/stdc++.h>
using namespace std;
long long int modpow(long long int a, long long int n, long long int temp) {
  long long int res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0)
    return (b);
  else
    return (gcd(b % a, a));
}
int main() {
  std::ios_base::sync_with_stdio(false);
  long long int a, b;
  cin >> a >> b;
  long long int res = 0;
  cout << b * (b - 1) / 2 % 1000000007 *
              ((a * (a + 1) / 2 % 1000000007 * b) % 1000000007 + a) % 1000000007
       << "\n";
  return 0;
}

### Prompt

Generate a CPP solution 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;
long long int modpow(long long int a, long long int n, long long int temp) {
  long long int res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0)
    return (b);
  else
    return (gcd(b % a, a));
}
int main() {
  std::ios_base::sync_with_stdio(false);
  long long int a, b;
  cin >> a >> b;
  long long int res = 0;
  cout << b * (b - 1) / 2 % 1000000007 *
              ((a * (a + 1) / 2 % 1000000007 * b) % 1000000007 + a) % 1000000007
       << "\n";
  return 0;
}

```