# D. Present

## Description

Catherine received an array of integers as a gift for March 8. Eventually she grew bored with it, and she started calculated various useless characteristics for it. She succeeded to do it for each one she came up with. But when she came up with another one — xor of all pairwise sums of elements in the array, she realized that she couldn't compute it for a very large array, thus she asked for your help. Can you do it? Formally, you need to compute

Here $x \oplus y$ is a bitwise XOR operation (i.e. x ^ y in many modern programming languages). You can read about it in Wikipedia: https://en.wikipedia.org/wiki/Exclusive_or#Bitwise_operation.

Input
The first line contains a single integer n ($2 \leq n \leq 400\,000$) — the number of integers in the array.

The second line contains integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^7$).

Output
Print a single integer — xor of all pairwise sums of integers in the given array.

Examples
input
2
1 2
output
3
input
3
1 2 3
output
2
Note
In the first sample case there is only one sum 1+2=3.

In the second sample case there are three sums: 1+2=3, 1+3=4, 2+3=5. In binary they are represented as $011_2 \oplus 100_2 \oplus 101_2 = 010_2$, thus the answer is 2.

($\oplus$)is the bitwise xor operation. To define $x \oplus y$, consider binary representations of integers x and y. We put the i-th bit of the result to be 1 when exactly one of the i-th bits of x and y is 1. Otherwise, the i-th bit of the result is put to be 0.For example, $0101_2 \, \oplus \, 0011_2 = 0110_2$

## Code

#include<bits/stdc++.h>
using namespace std;
const int maxn = 4e5+10;
int a[maxn],b[maxn];
typedef long long ll;
int main()
{
ios::sync_with_stdio(0);
int n;ll ans=0;
cin>>n;
for(int i=1;i<=n;i++)    cin>>a[i];
for(int k=0;k<30;k++){//因为最大也就是2e6，所以30位二进制位是足够的
int mod=1<<(k+1);
for(int i=1;i<=n;i++)    b[i]=a[i]%mod;
sort(b+1,b+n+1);
ll mid=0;
for(int i=1;i<=n;i++){
mid+=upper_bound(b+i+1,b+n+1,(1<<(k+1))-1-b[i])-lower_bound(b+i+1,b+n+1,(1<<k)-b[i]);//为了避免重复的对数，每次从第i位往后找
mid+=upper_bound(b+i+1,b+n+1,(1<<(k+2))-b[i])-lower_bound(b+i+1,b+n+1,(1<<k)+(1<<(k+1))-b[i]);//注意upper和lower的使用
}
if(mid&1)    ans+=(1<<k);//如果第k位是1的对数有奇数个，那么答案中的第k位就是1，即(1<<k)($$2^k$$)
}
cout<<ans<<endl;
return 0;
}