Monkey A lives on a tree, he always plays on this tree.
One day, monkey A learned about one of the bit-operations, xor. He was keen of this interesting operation and wanted to practise it at once.
Monkey A gave a value to each node on the tree. And he was curious about a problem.
The problem is how large the xor result of number x and one node value of label y can be, when giving you a non-negative integer x and a node label u indicates that node y is in the subtree whose root is u(y can be equal to u).
Can you help him?
Input
There are no more than 6 test cases.
For each test case there are two positive integers n and q, indicate that the tree has n nodes and you need to answer q queries.
Then two lines follow.
The first line contains n non-negative integers V1,V2,⋯,Vn, indicating the value of node i.
The second line contains n-1 non-negative integers F1,F2,⋯Fn−1, Fi means the father of node i+1.
And then q lines follow.
In the i-th line, there are two integers u and x, indicating that the node you pick should be in the subtree of u, and x has been described in the problem.
2≤n,q≤105
0≤Vi≤109
1≤Fi≤n, the root of the tree is node 1.
1≤u≤n,0≤x≤109
Output
For each query, just print an integer in a line indicating the largest result.
Sample Input
2 2
1 2
1
1 3
2 1
Sample Output
2
3
可持久化Trie ，启发式合并即可，
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 200005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-5
#define pll pair<ll,ll>
#define lson 2*x
#define rson 2*x+1
long long qupower(int a, int b) {
long long ans = 1;
while (b) {
if (b & 1)ans = ans * a;
b >>= 1;
a = a * a;
}
return ans;
}
inline int read() {
int an = 0, x = 1; char c = getchar();
while (c > '9' || c < '0') {
if (c == '-') {
x = -1;
}
c = getchar();
}
while (c >= '0'&&c <= '9') {
an = an * 10 + c - '0'; c = getchar();
}
return an * x;
}
struct ask {
int u, id;
};
struct Trie {
Trie *nxt[2];
Trie() {
nxt[0] = nxt[1] = NULL;
}
};
int ans[maxn], val[maxn], n, q;
vector<int>G[maxn];
vector<ask>Q[maxn];
Trie *root[maxn];
void Insert(Trie *rt, int x) {
int id;
for (int i = 30; i >= 0; i--) {
id = 0;
if (x&(1 << i))id = 1;
if (rt->nxt[id] == NULL) {
Trie *tmp = new Trie();
rt->nxt[id] = tmp;
}
rt = rt->nxt[id];
}
}
int query(Trie *rt, int x) {
int ret = 0, id;
for (int i = 30; i >= 0; i--) {
id = 0;
if (x&(1 << i))id = 1;
if (rt->nxt[id ^ 1] != NULL) {
ret |= (1 << i);
rt = rt->nxt[id ^ 1];
}
else {
rt = rt->nxt[id];
}
}
return ret;
}
Trie *merge(Trie *p, Trie *q) {
if (p == NULL)return q;
if (q == NULL)return p;
p->nxt[0] = merge(p->nxt[0], q->nxt[0]);
p->nxt[1] = merge(p->nxt[1], q->nxt[1]);
free(q);
return p;
}
void build(int u) {
int v;
root[u] = new Trie();
Insert(root[u], val[u]);
for (int i = 0; i < G[u].size(); i++) {
v = G[u][i];
build(v);
root[u] = merge(root[u], root[v]);
}
for (int i = 0; i < Q[u].size(); i++) {
ans[Q[u][i].id] = query(root[u], Q[u][i].u);
}
}
void det(Trie *rt) {
if (rt->nxt[0])det(rt->nxt[0]);
if (rt->nxt[1])det(rt->nxt[1]);
free(rt);
}
void sol() {
int u, x;
ask tmp;
for (int i = 1; i <= n; i++) {
G[i].clear();
Q[i].clear();
cin >> val[i];
}
for (int i = 2; i <= n; i++) {
cin >> x;
G[x].push_back(i);
}
for (int i = 1; i <= q; i++) {
cin >> u >> x;
tmp.u = x;
tmp.id = i;
Q[u].push_back(tmp);
}
build(1);
for (int i = 1; i <= q; i++)cout << ans[i] << endl;
det(root[1]);
}
int main() {
ios::sync_with_stdio(false);
while (cin >> n >> q) {
sol();
}
}