【CODEFORCES】 A. Dreamoon and Sums

A. Dreamoon and Sums

time limit per test

1.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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?

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).

Sample test(s)

input

1 1

output

0

input

2 2

output

8

Note

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
",ans%1000000007);
    return 0;
}