题目大意：[NOI2014]起床困难综合症加强版，出到树上，$m(m\leqslant10^5)$组询问，带修改，数的范围$<2^{64})$
题解：先想到树剖，转化为序列问题，线段树维护区间从左到右和从右到左$01$变化值，然后发现如果对每一位考虑，复杂度为$O(64n\log_2^2n)$，$0.5s$时限并过不去。考虑去掉$k$，可以使用位运算来$O(1)$合并区间。
注意树剖询问的时候要按顺序修改，不可以反过来。
卡点：无
C++ Code：
#include <algorithm>
#include <cstdio>
#include <cctype>
namespace __IO {int x, ch;inline int read() {while (isspace(ch = getchar())) ;for (x = ch & 15; isdigit(ch = getchar()); ) x = x * 10 + (ch & 15);return x;}unsigned long long X;inline unsigned long long readllu() {while (isspace(ch = getchar())) ;for (X = ch & 15; isdigit(ch = getchar()); ) X = X * 10 + (ch & 15);return X;}
}
using __IO::read;
using __IO::readllu;const int maxn = 100010;
const unsigned long long inf = ~0;int Opt[maxn], __Opt[maxn];
unsigned long long Num[maxn], __Num[maxn];
int n, m, __p;
inline void calc(unsigned long long &x, int y, unsigned long long z) {switch (y) {case 1: x &= z; break;case 2: x |= z; break;case 3: x ^= z;}
}
struct node {unsigned long long __0, __1;inline friend node operator + (const node &lhs, const node &rhs) {return (node) {(lhs.__0 & rhs.__1) | (~lhs.__0 & rhs.__0),(lhs.__1 & rhs.__1) | (~lhs.__1 & rhs.__0)};}
} ;
namespace SgT {const int N = maxn << 2;node lr[N], rl[N];void build(const int rt, const int l, const int r) {if (l == r) {calc(lr[rt].__0 = 0, Opt[l], Num[l]);calc(lr[rt].__1 = inf, Opt[l], Num[l]);calc(rl[rt].__0 = 0, Opt[l], Num[l]);calc(rl[rt].__1 = inf, Opt[l], Num[l]);return ;}const int mid = l + r >> 1, lc = rt << 1, rc = rt << 1 | 1;build(lc, l, mid), build(rc, mid + 1, r);lr[rt] = lr[lc] + lr[rc];rl[rt] = rl[rc] + rl[lc];}int pos;void __modify(const int rt, const int l, const int r) {if (l == r) {calc(lr[rt].__0 = 0, Opt[l], Num[l]);calc(lr[rt].__1 = inf, Opt[l], Num[l]);calc(rl[rt].__0 = 0, Opt[l], Num[l]);calc(rl[rt].__1 = inf, Opt[l], Num[l]);return ;}const int mid = l + r >> 1, lc = rt << 1, rc = rt << 1 | 1;if (pos <= mid) __modify(lc, l, mid);else __modify(rc, mid + 1, r);lr[rt] = lr[lc] + lr[rc];rl[rt] = rl[rc] + rl[lc];}void modify(int __pos, int y, unsigned long long z) {pos = __pos;Opt[pos] = y;Num[pos] = z;__modify(1, 1, n);}int L, R;node res;void querylr(const int rt, const int l, const int r) {if (L <= l && R >= r) {res = res + lr[rt];return ;}const int mid = l + r >> 1;if (L <= mid) querylr(rt << 1, l, mid);if (R > mid) querylr(rt << 1 | 1, mid + 1, r);}void queryrl(const int rt, const int l, const int r) {if (L <= l && R >= r) {res = res + rl[rt];return ;}const int mid = l + r >> 1;if (R > mid) queryrl(rt << 1 | 1, mid + 1, r);if (L <= mid) queryrl(rt << 1, l, mid);}node query(int __L, int __R) {res.__0 = 0, res.__1 = inf;if (__L <= __R) {L = __L, R = __R;querylr(1, 1, n);} else {L = __R, R = __L;queryrl(1, 1, n);}return res;}
}int head[maxn], cnt;
struct Edge {int to, nxt;
} e[maxn << 1];
inline void addedge(int a, int b) {e[++cnt] = (Edge) { b, head[a] }; head[a] = cnt;e[++cnt] = (Edge) { a, head[b] }; head[b] = cnt;
}int fa[maxn], dep[maxn], sz[maxn];
int dfn[maxn], idx, top[maxn], son[maxn];
void dfs1(int u) {sz[u] = 1;for (int i = head[u]; i; i = e[i].nxt) {int v = e[i].to;if (v != fa[u]) {fa[v] = u;dep[v] = dep[u] + 1;dfs1(v);if (!son[u] || sz[v] > sz[son[u]]) son[u] = v;sz[u] += sz[v];}}
}
void dfs2(int u) {dfn[u] = ++idx;int v = son[u];if (v) top[v] = top[u], dfs2(v);for (int i = head[u]; i; i = e[i].nxt) {int v = e[i].to;if (v != fa[u] && v != son[u]) {top[v] = v;dfs2(v);}}
}unsigned long long query(int x, int y, unsigned long long z) {static node res, S[maxn];int tot = 0;res.__0 = 0, res.__1 = inf;while (top[x] != top[y]) {if (dep[top[x]] > dep[top[y]]) {res = res + SgT::query(dfn[x], dfn[top[x]]);x = fa[top[x]];} else {S[++tot] = SgT::query(dfn[top[y]], dfn[y]);y = fa[top[y]];}}res = res + SgT::query(dfn[x], dfn[y]);while (tot) res = res + S[tot--];unsigned long long ans = 0, ret = 0;for (int i = 63; ~i; --i) {if (res.__0 >> i & 1) ret |= 1ull << i;else if (res.__1 >> i & 1) {if ((ans | 1ull << i) <= z) {ans |= 1ull << i;ret |= 1ull << i;}}}return ret;
}int main() {n = read(), m = read(), __p = read();for (int i = 1; i <= n; ++i) {__Opt[i] = read(), __Num[i] = readllu();}for (int i = 1, a, b; i < n; ++i) {a = read(), b = read();addedge(a, b);}dfs1(1), dfs2(top[1] = 1);for (int i = 1; i <= n; ++i) Opt[dfn[i]] = __Opt[i], Num[dfn[i]] = __Num[i];SgT::build(1, 1, n);while (m --> 0) {static int op, x, y;static unsigned long long z;op = read(), x = read(), y = read(), z = readllu();if (op == 1) {printf("%llu\n", query(x, y, z));} else {SgT::modify(dfn[x], y, z);}}return 0;
}