#include <bits/stdc++.h>
using namespace std;
long long a, b;
long long pow_mod(long long x, int yu) {
  long long res = 1;
  while (yu) {
    if (yu & 1) {
      res *= x;
      res %= 1000000007;
    }
    x *= x;
    x %= 1000000007;
    yu >>= 1;
  }
  return res;
}
int main() {
  scanf("%I64d%I64d", &a, &b);
  long long ans = 0;
  long long pp = pow_mod(2, 1000000007 - 2);
  for (int i = 1; i < b; i++) {
    long long c = a * (a + 1);
    c %= 1000000007;
    c *= pp;
    c %= 1000000007;
    c *= i;
    c %= 1000000007;
    c *= b;
    c %= 1000000007;
    long long d = a * i;
    d %= 1000000007;
    ans += c + d;
    ans %= 1000000007;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
long long a, b;
long long pow_mod(long long x, int yu) {
  long long res = 1;
  while (yu) {
    if (yu & 1) {
      res *= x;
      res %= 1000000007;
    }
    x *= x;
    x %= 1000000007;
    yu >>= 1;
  }
  return res;
}
int main() {
  scanf("%I64d%I64d", &a, &b);
  long long ans = 0;
  long long pp = pow_mod(2, 1000000007 - 2);
  for (int i = 1; i < b; i++) {
    long long c = a * (a + 1);
    c %= 1000000007;
    c *= pp;
    c %= 1000000007;
    c *= i;
    c %= 1000000007;
    c *= b;
    c %= 1000000007;
    long long d = a * i;
    d %= 1000000007;
    ans += c + d;
    ans %= 1000000007;
  }
  printf("%I64d\n", ans);
  return 0;
}

```