注意到对于树边 $(x, y)$，$d(y)$ 表示根节点到 $y$ 的距离，那么有
$d(y) = d(x) \oplus e(x, y)$，由于异或操作对于路径重复的部分，值为 $0$
所以 $\oplus (x \to y)$ 实际上就是 $d(x) \oplus d(y)$
只要通过 $\text{dfs}$ 预处理出所有的 $d[x]$，然后转换成最大异或对问题求解即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
#include <cmath>
#include <bitset>
#include <assert.h>
#include <unordered_map>

using namespace std;
typedef long long ll;

#define Cmp(a, b) memcmp(a, b, sizeof(b))
#define Cpy(a, b) memcpy(a, b, sizeof(b))
#define Set(a, v) memset(a, v, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define _forS(i, l, r) for(set<int>::iterator i = (l); i != (r); i++)
#define _rep(i, l, r) for(int i = (l); i <= (r); i++)
#define _for(i, l, r) for(int i = (l); i < (r); i++)
#define _forDown(i, l, r) for(int i = (l); i >= r; i--)
#define debug_(ch, i) printf(#ch"[%d]: %d\n", i, ch[i])
#define debug_m(mp, p) printf(#mp"[%d]: %d\n", p->first, p->second)
#define debugS(str) cout << "dbg: " << str << endl;
#define debugArr(arr, x, y) _for(i, 0, x) { _for(j, 0, y) printf("%c", arr[i][j]); printf("\n"); }
#define _forPlus(i, l, d, r) for(int i = (l); i + d < (r); i++)
#define lowbit(i) (i & (-i))
#define MPR(a, b) make_pair(a, b)

pair<int, int> crack(int n) {
    int st = sqrt(n);
    int fac = n / st;
    while (n % st) {
        st += 1;
        fac = n / st;
    }

    return make_pair(st, fac);
}

inline ll qpow(ll a, int n) {
    ll ans = 1;
    for(; n; n >>= 1) {
        if(n & 1) ans *= 1ll * a;
        a *= a;
    }
    return ans;
}

template <class T>
inline bool chmax(T& a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a % b);
}

ll ksc(ll a, ll b, ll mod) {
    ll ans = 0;
    for(; b; b >>= 1) {
        if (b & 1) ans = (ans + a) % mod;
        a = (a * 2) % mod;
    }
    return ans;
}

ll ksm(ll a, ll b, ll mod) {
    ll ans = 1 % mod;
    a %= mod;

    for(; b; b >>= 1) {
        if (b & 1) ans = ksc(ans, a, mod);
        a = ksc(a, a, mod);
    }

    return ans;
}

template <class T>
inline bool chmin(T& a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool lexSmaller(vector<T> a, vector<T> b) {
    int n = a.size(), m = b.size();
    int i;
    for(i = 0; i < n && i < m; i++) {
        if (a[i] < b[i]) return true;
        else if (b[i] < a[i]) return false;
    }
    return (i == n && i < m);
}

// ============================================================== //

const int maxn = 3100000 + 5;

class Graph {
public:
    int n;
    int tot = 1;
    vector<int> head, e, ne, ver;

    Graph() = default;
    Graph(int _n) : n(_n) {
        tot = 1;
        head.resize(n), e.resize(n<<1), ne.resize(n<<1), ver.resize(n<<1);
    }

    void add(int a, int b, int c) {
        ver[++tot] = b; e[tot] = c; ne[tot] = head[a]; head[a] = tot;
    }
};

const int N = 100000 + 5;
int n, d[N];
Graph g(N);

class Trie {
public:
    int t[maxn][2];
    int tot;

    Trie() {
        tot = 1;
        memset(t, 0, sizeof t);
    }

    void insert(int x) {
        int p = 1;
        for (int i = 30; i >= 0; i--) {
            int c = x >> i & 1;
            if (!t[p][c]) t[p][c] = ++tot;
            p = t[p][c];
        }
    }

    int query(int x) {
        int res = 0, p = 1;
        for (int i = 30; i >= 0; i--) {
            int c = x >> i & 1;
            if (t[p][!c]) p = t[p][!c], res += (1<<i);
            else p = t[p][c];
        }
        return res;
    }
} trie;

void dfs(int x, int fa) {
    for (int i = g.head[x]; i; i = g.ne[i]) {
        int y = g.ver[i];
        if (y == fa) continue;

        d[y] = d[x] ^ g.e[i];
        dfs(y, x);
    }
}

int main() {
    //freopen("input.txt", "r", stdin);
    memset(d, 0, sizeof d);
    scanf("%d", &n);
    for (int i = 0; i < n-1; i++) {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        a++, b++;
        g.add(a, b, c), g.add(b, a, c);
    }

    // dfs
    dfs(1, 0);

    // trie
    for (int i = 1; i <= n; i++) trie.insert(d[i]);
    int res = 0;
    for (int i = 1; i <= n; i++) res = max(res, trie.query(d[i]));
    printf("%d\n", res);
}