#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
long long mod(long long x) {
  x %= MOD;
  if (x < 0) x += MOD;
  return x;
}
const bool D = 0;
int main() {
  if (D) {
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
  } else {
  }
  int a;
  long long b;
  scanf("%d%I64d", &a, &b);
  long long t = b - 1;
  t = mod(t * (t + 1) / 2);
  long long ans = 0;
  for (int k = 1; k <= a; ++k) {
    ans = mod(ans + mod(t * mod(k * b + 1)));
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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 long long MOD = 1000000007;
long long mod(long long x) {
  x %= MOD;
  if (x < 0) x += MOD;
  return x;
}
const bool D = 0;
int main() {
  if (D) {
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
  } else {
  }
  int a;
  long long b;
  scanf("%d%I64d", &a, &b);
  long long t = b - 1;
  t = mod(t * (t + 1) / 2);
  long long ans = 0;
  for (int k = 1; k <= a; ++k) {
    ans = mod(ans + mod(t * mod(k * b + 1)));
  }
  printf("%I64d\n", ans);
  return 0;
}

```