#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-8;
const long long mod = 1000000007LL;
const long long inv = 500000004LL;
long long mult(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long sub(long long a, long long b) {
  return (mod * 2LL + (a % mod) - (b % mod)) % mod;
}
long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long a, b;
  cin >> a >> b;
  long long ans = 0;
  long long as = a * (a + 1) / 2LL;
  long long all = mult(as, b);
  all = add(a, all);
  for (long long c = 1; c < b; c++) {
    long long tp = mult(c, all);
    ans = add(ans, tp);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate 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;
const double pi = acos(-1.0);
const double eps = 1e-8;
const long long mod = 1000000007LL;
const long long inv = 500000004LL;
long long mult(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long sub(long long a, long long b) {
  return (mod * 2LL + (a % mod) - (b % mod)) % mod;
}
long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long a, b;
  cin >> a >> b;
  long long ans = 0;
  long long as = a * (a + 1) / 2LL;
  long long all = mult(as, b);
  all = add(a, all);
  for (long long c = 1; c < b; c++) {
    long long tp = mult(c, all);
    ans = add(ans, tp);
  }
  cout << ans << endl;
  return 0;
}

```