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 MAX_BUF_SIZE = 16384;
char BUFOR[MAX_BUF_SIZE];
int BUF_SIZE, BUF_POS;
char ZZZ;
int _MINUS;
const int MXN = 2000010;
const int C = 262144;
const int INF = 1000000001;
const int MOD = 1000000007;
int a, b;
void test() {
  scanf("%d %d", &a, &b);
  long long int x = ((long long int)b * (b - 1) / 2) % MOD;
  long long int y = ((long long int)(a + 1) * a / 2) % MOD;
  y = ((long long int)y * b) % MOD;
  y = (a + y) % MOD;
  printf("%d\n", (x * y) % MOD);
}
int main() {
  int te = 1;
  while (te--) test();
  return 0;
}
