如果是求和就很好做了...
不是求和也无伤大雅....
一维太难限制条件了，考虑二维限制
一维$dfs$序，一维$dep$序
询问$(x, k)$对应着在$dfs$上查$[dfn[x], dfn[x] + sz[x] - 1]$，在$dep$序上查$[dep[x], dep[x] + k]$
这样子，每个询问对应查询一段矩形内的最小值
然而树套树是过不了的.....
发现一个询问看似在$dep$序上对应了一段区间，实际上可以扩展到对应一段前缀
这样子，只需要一个主席树就可以做到了
复杂度$O(n \log n)$
#include <map>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
namespace remoon {
#define ri register int
#define tpr template <typename ra>
#define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++)
#define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --)
#define gc getchar
inline int read() {
int p = , w = ; char c = gc();
while(c > '' || c < '') { if(c == '-') w = -; c = gc(); }
while(c >= '' && c <= '') p = p * + c - '', c = gc();
return p * w;
}
int wr[], rw;
#define pc(iw) putchar(iw)
tpr inline void write(ra o, char c = '\n') {
if(!o) pc('');
if(o < ) o = -o, pc('-');
while(o) wr[++ rw] = o % , o /= ;
while(rw) pc(wr[rw --] + '');
pc(c);
}
tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; }
tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; }
tpr inline bool ckmin(ra &a, ra b) { return (a > b) ? a = b, : ; }
tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b, : ; }
}
using namespace std;
using namespace remoon; #define sid 300050
#define oid 12050000 int dfn[sid], sz[sid];
int n, r, m, id, tim, cnp, mxd;
int rt[sid], nxt[sid], node[sid], cap[sid];
int fa[sid], q[sid], w[sid], dep[sid];
int ls[oid], rs[oid], miv[oid]; inline void addedge(int u, int v) {
nxt[++ cnp] = cap[u]; cap[u] = cnp; node[cnp] = v;
} #define cur node[i]
inline void dfs(int o, int f) {
fa[o] = f; dep[o] = dep[f] + ;
sz[o] = ; dfn[o] = ++ tim;
for(int i = cap[o]; i; i = nxt[i])
if(cur != f) dfs(cur, o), sz[o] += sz[cur];
} inline void insert(int &now, int pre, int l, int r, int p, int v) {
now = ++ id;
ls[now] = ls[pre]; rs[now] = rs[pre];
miv[now] = min(miv[pre], v);
if(l == r) return;
int mid = (l + r) >> ;
if(p <= mid) insert(ls[now], ls[pre], l, mid, p, v);
else insert(rs[now], rs[pre], mid + , r, p, v);
} inline void build() {
int fr = , to = ;
q[++ to] = r; miv[] = 1e9;
while(fr <= to) {
int o = q[fr];
for(ri i = cap[o]; i; i = nxt[i])
if(cur != fa[o]) q[++ to] = cur;
if(dep[o] != dep[q[fr - ]])
insert(rt[dep[o]], rt[dep[o] - ], , n, dfn[o], w[o]);
else insert(rt[dep[o]], rt[dep[o]], , n, dfn[o], w[o]);
fr ++; cmax(mxd, dep[o]);
}
} inline int qry(int o, int l, int r, int ml, int mr) {
if(ml > r || mr < l || !o) return 1e9;
if(ml <= l && mr >= r) return miv[o];
int mid = (l + r) >> ;
return min(qry(ls[o], l, mid, ml, mr), qry(rs[o], mid + , r, ml, mr));
} int main() {
n = read(); r = read();
rep(i, , n) w[i] = read();
rep(i, , n) {
int u = read(), v = read();
addedge(u, v); addedge(v, u);
}
dfs(r, ); build();
int lst = ; m = read();
rep(i, , m) {
int x = (read() + lst) % n + ;
int k = (read() + lst) % n;
write(lst = qry(rt[min(dep[x] + k, mxd)], , n, dfn[x], dfn[x] + sz[x] - ));
}
return ;
}