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  and ,
where k is some integer number in range[1, a].

By  we
denote the quotient of integer division of x and y.
By  we
denote the remainder of integer division of x andy.
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?

The single line of the input contains two integers a, b (1 ≤ a, b ≤ 107).

Print a single integer representing the answer modulo 1 000 000 007 (109 + 7).

For the first sample, there are no nice integers because  is
always zero.

For the second sample, the set of nice integers is {3, 5}.

题解：这一题能够将题目中的那个东西化简，最后变成x=（k*b+1）*(x mod b)他要求全部x的和，由于k是1~a，x mod b是1~b-1，所以我们能够直接把这个结果算出来。简单的求和就可以。

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

long long a,b;

int main()

{

    scanf("%I64d%I64d",&a,&b);

    long long c=((b-1)*b/2)%1000000007,d=((b*(a*(a+1)/2%1000000007)+a)%1000000007);

    long long ans=(c*d)%1000000007;

    printf("%I64d\n",ans%1000000007);

    return 0;

}