codeforces 476c Dreamoon and Sums

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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}.

给你此公式 ，k的范围是[1,a],然后给你a和b，让你求满足此条件的x的和。

此公式可以化简为：x=k*mod(x,b)*b+mod(x,b) 即x=(k*b+1)*mod(x,b)，k可以枚举，b已知，而mod(x,b)的取值是1~b，可以通过公式直接求出。

//171 ms 100 KB#include<stdio.h>#include<string.h>#define ll long long#define mod 1000000007int main(){    ll a,b;    while(scanf("%I64d%I64d",&a,&b)!=EOF)    {        ll ans=0;        for(ll i=1;i<=a;i++)        {            ll c=(i*b+1)%mod;            ll d=((b-1)*b/2)%mod;            ans=(ans+c*d)%mod;        }        printf("%I64d\n",ans);    }    return 0;}