It is well known that the distance between two points xy on the tree is deep[x]+deep[y]-deep[lca(x,y)]*2
Then we replace this addition and subtraction with XOR, we will find that deep[lca(x,y)] is eliminated
So the problem is simplified to w is the prefix XOR sum of each point, as long as you find a pair of the largest (x, y) to make w[x]^w[y] the largest, this classic problem can be solved with a dictionary tree . Be careful not to overdose memset
/// .-~~~~~~~~~-._ _.-~~~~~~~~~-.
/// __.' ~. .~ `.__
/// .'// \./ \\`.
/// .'// | \\`.
/// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`.
/// .'//.-" `-. | .-' "-.\\`.
/// .'//______.============-.. \ | / ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma GCC optimize("Ofast")
#pragma comment(linker, "/STACK:102400000,102400000")
#pragma GCC target(sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx)
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <fstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;
#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
#define sz size
#define fi first
#define se second
#define pf printf
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e2+7;
const int _=1e5+10;
//const int maxx=1e7+10;
const double EPS=1e-8;
const double eps=1e-8;
const LL mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}
inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}
void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;
int n;
int Next[_*32][2];
int val[_*32];
int st;
void init() {
me(Next[0],0);
me(val,0);
st=1;
}
void insert(LL x) {
int u=0;
for(int i=32;i>=0;i--) {
int c=((x>>i)&1);
if(!Next[u][c]) {
me(Next[st],0);
Next[u][c]=st++;
}
u=Next[u][c];
++val[u];
}
}
LL query(LL x) {
LL ans=0;
int t=0;
for(int i=32;i>=0;i--) {
int k=((x>>i)&1);
if(Next[t][k^1]) {
t=Next[t][k^1];
ans|=1<<i;
}
else t=Next[t][k];
}
return ans;
}
int fa[_],w[_];
void solve() {
s_1(n);
init();
insert(0);
LL ans=0;
for(int i=2;i<=n;i++) {
s_2(fa[i],w[i]);
w[i]^=w[fa[i]];
ans=max(ans,query(w[i]));
insert(w[i]);
}
print(ans);
}
int main() {
//freopen( "1.in" , "r" , stdin );
//freopen( "1.out" , "w" , stdout );
int t=1;
//init();
s_1(t);
for(int cas=1;cas<=t;cas++) {
//printf("Case #%d: ",cas);
solve();
}
}