#include <bits/stdc++.h>
using namespace std;
const int INF = 1000 * 1000 * 1000 + 5;
const int MOD = 1000 * 1000 * 1000 + 7;
const long double EPS = 1e-9;
const int N = 1000 * 10000 + 5;
long long a, b;
long long f[N];
long long ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> a >> b;
  f[1] = 1;
  for (int i = 2; i <= a; ++i) f[i] = (f[i - 1] + i) % MOD;
  for (int i = 1; i <= b - 1; ++i)
    ans = (ans + (((i * f[a]) % MOD) * b) % MOD + (i * a) % MOD) % MOD;
  cout << ans << '\n';
  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>
using namespace std;
const int INF = 1000 * 1000 * 1000 + 5;
const int MOD = 1000 * 1000 * 1000 + 7;
const long double EPS = 1e-9;
const int N = 1000 * 10000 + 5;
long long a, b;
long long f[N];
long long ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> a >> b;
  f[1] = 1;
  for (int i = 2; i <= a; ++i) f[i] = (f[i - 1] + i) % MOD;
  for (int i = 1; i <= b - 1; ++i)
    ans = (ans + (((i * f[a]) % MOD) * b) % MOD + (i * a) % MOD) % MOD;
  cout << ans << '\n';
  return 0;
}

```