#include <bits/stdc++.h>
#pragma comment(linker, "/stack:16777216")
using namespace std;
const int INF = 1000000000;
const int MAX = 500;
const int MAX2 = 3007;
const int ADD = 1502;
const int BASE = 1000000000;
const int MOD = 1000000007;
long long a, b;
int main() {
  cin >> a >> b;
  long long res = 0;
  for (int m = (1); m < (b); ++m) {
    long long cnt = a;
    long long sum = (a + b * (((a) * (a + 1) / 2) % MOD)) % MOD * m;
    sum %= MOD;
    res += sum;
    res %= MOD;
  }
  cout << res << 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>
#pragma comment(linker, "/stack:16777216")
using namespace std;
const int INF = 1000000000;
const int MAX = 500;
const int MAX2 = 3007;
const int ADD = 1502;
const int BASE = 1000000000;
const int MOD = 1000000007;
long long a, b;
int main() {
  cin >> a >> b;
  long long res = 0;
  for (int m = (1); m < (b); ++m) {
    long long cnt = a;
    long long sum = (a + b * (((a) * (a + 1) / 2) % MOD)) % MOD * m;
    sum %= MOD;
    res += sum;
    res %= MOD;
  }
  cout << res << endl;
  return 0;
}

```