#include <bits/stdc++.h>
const int N = (int)1e6 + 123;
const double eps = 1e-6;
const long long inf = (long long)1e18 + 123;
const int mod = (int)1e9 + 7;
using namespace std;
long long a, b;
long long ans;
int main() {
  cin >> a >> b;
  long long sum = 1ll * a * (a + 1) / 2 % mod;
  for (long long i = 1; i < b; i++) {
    ans = (ans + 1ll * i * a % mod) % mod;
    ans = (ans +
           1ll * (1ll * b * (1ll * a * (a + 1) / 2 % mod) % mod) * i % mod) %
          mod;
  }
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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>
const int N = (int)1e6 + 123;
const double eps = 1e-6;
const long long inf = (long long)1e18 + 123;
const int mod = (int)1e9 + 7;
using namespace std;
long long a, b;
long long ans;
int main() {
  cin >> a >> b;
  long long sum = 1ll * a * (a + 1) / 2 % mod;
  for (long long i = 1; i < b; i++) {
    ans = (ans + 1ll * i * a % mod) % mod;
    ans = (ans +
           1ll * (1ll * b * (1ll * a * (a + 1) / 2 % mod) % mod) * i % mod) %
          mod;
  }
  cout << ans;
  return 0;
}

```