CF_476C_DreamoonAndSums

C. 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}.

本来以为是数论题

后来仔细一看呵呵……

简单的等差数列求和就可以了

给定a和b，那么其实满足条件的数就是下面这些数(列写下就好了)

b+1   2b+1   3b+1   …………   ab+1

2b+2  4b+2  6b+2  …………  2ab+2

3b+3 6b+3  9b+3   ………… 3ab +3

b(b-1)+(b-1)  2(b-1)b+(b-1)  3(b-1)b+(b-1) …………a(b-1)b+(b-1)

横着每行a个数，一共b-1行 求和就可以了

#include <iostream>#include <stdio.h>using namespace std;typedef long long LL;const int MO=1e9+7;int main(){    LL a,b;    scanf("%I64d %I64d",&a,&b);    LL ans1=(b*(b-1)/2)%MO;    LL ans2=((((a*(a+1))/2)%MO*b)%MO*ans1)%MO;    LL ans=((ans1*a)%MO+ans2)%MO;    printf("%I64d\n",ans);    return 0;}