#include <bits/stdc++.h>
using namespace std;
const long long M = 1000000007;
long long i, j, k, l, m, n, o, p, ans, cul, a, b, r;
int main() {
  cin >> a >> b;
  ans = 0;
  for (r = 1; r <= b - 1; r++) {
    o = a + 1;
    p = a;
    if (o & 1)
      p >>= 1;
    else
      o >>= 1;
    cul = (o * p) % M;
    cul = (cul * b) % M;
    cul += a;
    if (cul >= M) cul -= M;
    ans += (cul * r) % M;
    if (ans >= M) ans -= M;
  }
  cout << ans << 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>
using namespace std;
const long long M = 1000000007;
long long i, j, k, l, m, n, o, p, ans, cul, a, b, r;
int main() {
  cin >> a >> b;
  ans = 0;
  for (r = 1; r <= b - 1; r++) {
    o = a + 1;
    p = a;
    if (o & 1)
      p >>= 1;
    else
      o >>= 1;
    cul = (o * p) % M;
    cul = (cul * b) % M;
    cul += a;
    if (cul >= M) cul -= M;
    ans += (cul * r) % M;
    if (ans >= M) ans -= M;
  }
  cout << ans << endl;
  return 0;
}

```