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}.
#include <bits/stdc++.h>
using namespace std;
const int inf = (1 << 29);
const long long MOD = 1000000007LL;
int a, b;
long long GAUSS;
void citeste() { scanf("%d %d", &a, &b); }
long long gauss(int a) {
  long long y = (long long)a;
  y = (y * (y + 1LL)) / 2LL;
  y %= MOD;
  return y;
}
long long formula(long long x) {
  long long ans = (x * GAUSS) % MOD;
  return ans;
}
void rezolva() {
  long long ans = 0LL;
  GAUSS = gauss(b - 1);
  for (int k = 1; k <= a; ++k) {
    long long y = (1LL * b * k + 1LL) % MOD;
    ans += formula(y);
    ans %= MOD;
  }
  cout << ans << "\n";
}
int main() {
  citeste();
  rezolva();
  return 0;
}
