#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long pw(long long a, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1ll) ret = ret * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ret;
}
inline void solve() {
  int a, b;
  cin >> a >> b;
  long long q = b * 1ll * (b - 1) % mod;
  q = q * pw(2, mod - 2) % mod;
  long long qa = a * 1ll * (a + 1) % mod;
  qa = qa * pw(2, mod - 2) % mod;
  qa = qa * b % mod;
  long long ans = q * (a + qa) % mod;
  cout << ans << endl;
}
int main() {
  int t = 1;
  for (int i = 0; i < t; ++i) {
    solve();
  }
  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;
long long mod = 1000000007;
long long pw(long long a, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1ll) ret = ret * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ret;
}
inline void solve() {
  int a, b;
  cin >> a >> b;
  long long q = b * 1ll * (b - 1) % mod;
  q = q * pw(2, mod - 2) % mod;
  long long qa = a * 1ll * (a + 1) % mod;
  qa = qa * pw(2, mod - 2) % mod;
  qa = qa * b % mod;
  long long ans = q * (a + qa) % mod;
  cout << ans << endl;
}
int main() {
  int t = 1;
  for (int i = 0; i < t; ++i) {
    solve();
  }
  return 0;
}

```