codeforces 477A A. Dreamoon and Sums(数学)

题目链接：

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 $\text{[math]}$ and $\text{[math]}$ , where k is some integer number in range[1, a].

By $\text{[math]}$ we denote the quotient of integer division of x and y. By $\text{[math]}$ 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).

Examples

input

1 1

output

input

2 2

output

8

题意：

计算满足上述要求的数的和,满足要求的数x/b是x%b的k倍,k属于[1,a];

思路：

一个推公式的题,x=b*div+mod;其中div=k*mod;所以x=(b*k+1)*mod;而1<=k<=a;1<=mod<=b-1;
得到公式∑∑(b*k+1)*mod=(∑(b*k+1))*(∑mod)=b*(b-1)/2*(a+a*(a+1)/2*b);
就变成水题了;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
  
using namespace std;
  
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
  
typedef  long long LL;
  
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
  
const int mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e6+20;
const int maxn=1e3+110;
const double eps=1e-12;
 

int main()
{
    LL a,b;
    read(a);read(b);
    LL ans=b*(b-1)/2%mod;
    ans=ans*(a+a*(a+1)/2%mod*b%mod)%mod;
    cout<<ans<<endl;
    
    return 0;
}