[Codeforces Round #626 (Div. 2, based on Moscow Open Olympiad in Informatics)] -D. Present（异或性质，按位拆分，树桩数组）

D. Present

time limit per test

3 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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

(a1+a2)⊕(a1+a3)⊕…⊕(a1+an)⊕(a2+a3)⊕…⊕(a2+an)…⊕(an−1+an)(a1+a2)⊕(a1+a3)⊕…⊕(a1+an)⊕(a2+a3)⊕…⊕(a2+an)…⊕(an−1+an)

Here x⊕yx⊕y is a bitwise XOR operation (i.e. xx ^ yy 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 nn (2≤n≤4000002≤n≤400000) — the number of integers in the array.

The second line contains integers a1,a2,…,ana1,a2,…,an (1≤ai≤1071≤ai≤107).

Output

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

Examples

input

Copy

2
1 2

output

Copy

3

input

Copy

3
1 2 3

output

Copy

2

Note

In the first sample case there is only one sum 1+2=31+2=3.

In the second sample case there are three sums: 1+2=31+2=3, 1+3=41+3=4, 2+3=52+3=5. In binary they are represented as 0112⊕1002⊕1012=01020112⊕1002⊕1012=0102, thus the answer is 2.

⊕⊕ is the bitwise xor operation. To define x⊕yx⊕y, consider binary representations of integers xx and yy. We put the ii-th bit of the result to be 1 when exactly one of the ii-th bits of xx and yy is 1. Otherwise, the ii-th bit of the result is put to be 0. For example, 01012⊕00112=0110201012⊕00112=01102.

题意：

给定一个含有n个整数的数组，让你求

\((a_1 + a_2) \oplus (a_1 + a_3) \oplus \ldots \oplus (a_1 + a_n) \\ \oplus (a_2 + a_3) \oplus \ldots \oplus (a_2 + a_n) \\ \ldots \\ \oplus (a_{n-1} + a_n) \\\)

思路：

我们知道异或是不作进位的加法，所以我们考虑计算答案的每一位，从而得出答案值。

对于第\(\mathit k\)位（从第0位开始），我们考虑将所有的\(a_i\)对\(2^{k+1}\)取模后有多少对\((i,j)\)使其\(sum=a_i+a_j\)的第\(\mathit k\)位为1，

我们知道在取模操作后第k位为1的话，sum要在\([2^k, 2^{k+1}),[2^{k+1} + 2^k, 2^{k+2} - 2]\) 这2个区间中。

可以将取模后的数字取模后用转指针或者二分法来求点对的个数，

这里我用的是树状数组记录数值出现的次数来求得，时间复杂度和空间复杂度都不优，仅做一种可行写法示例。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#include <sstream>
#include <bitset>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
const int maxn = 100000100;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int tree[maxn];
int lowbit(int x)
{
    return (-x)& x;
}
void add(int pos, int val)
{
    pos++;
    while (pos < maxn)
    {
        tree[pos] += val;
        pos += lowbit(pos);
    }
}
int ask(int pos)
{
    pos++;
    int res = 0;
    while (pos > 0)
    {
        res += tree[pos];
        pos -= lowbit(pos);
    }
    return res;
}

int n;
int a[400010];
int query(int l, int r)
{
    return ask(r) - ask(l - 1);
}
int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    n = readint();
    repd(i, 1, n)
    {
        a[i] = readint();
    }
    ll ans = 0ll;
    for (int i = 0; i <= 24; ++i)
    {
        ll res = 0ll;
        repd(j, 1, n)
        {
            int x = a[j] % (1 << i + 1);
            res += query((1 << i) - x, (1 << i + 1) - 1 - x);
            res += query((1 << i) + (1 << i + 1) - x, (1 << i + 2) - 2 - x);
            add(x, 1);
        }
        repd(j, 1, n)
        {
            int x = a[j] % (1 << i + 1);
            add(x, -1);
        }
        if (res & 1)
        {
            ans += (1 << i);
        }
    }
    printf("%lld\n", ans );
    return 0;
}