取余

题目网址： http://codeforces.com/contest/424/problem/C

Description

People in the Tomskaya region like magic formulas very much. You can see some of them below.

Imagine you are given a sequence of positive integer numbers p₁, p₂, ..., p_n. Lets write down some magic formulas:

Here, "mod" means the operation of taking the residue after dividing.

The expression  means applying the bitwise xor (excluding "OR") operation to integers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by "^", in Pascal — by "xor".

People in the Tomskaya region like magic formulas very much, but they don't like to calculate them! Therefore you are given the sequence p, calculate the value of Q.

Input

The first line of the input contains the only integer n (1 ≤ n ≤ 10⁶). The next line contains n integers: p₁, p₂, ..., p_n (0 ≤ p_i ≤ 2·10⁹).

Output

The only line of output should contain a single integer — the value of Q.

Sample Input

Input

3
1 2 3

Output

3

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstdio>
using namespace std;
long long a[1000005];

int main()
{
    int i,n;
    long long p,sum;
    a[0]=0;
    for(i=1;i<=1000004;i++)
    a[i]=a[i-1]^i;
    while(cin>>n)
    {
       sum=0;
       for(i=0;i<n;i++)
       {
           scanf("%lld",&p);
           sum=sum^p;
       }
       for(i=2;i<=n;i++)
       {
           int c=n%i;
           int cc=n/i;
           if(cc&1)
           {
               sum=sum^a[i-1];
           }
           sum=sum^a[c];
       }
        cout<<sum<<endl;
    }
    return 0;
}