给定长度为n的数列X={x1,x2,...,xn}和长度为m的数列Y={y1,y2,...,ym}，令矩阵A中第i行第j列的值Aij=xi xor  yj，每次询问给定矩形区域i∈[u,d],j∈[l,r]，找出第k大的Aij。

第一行包含两个正整数n,m，分别表示两个数列的长度

共p行，每行包含一个非负整数，表示此次询问的答案。

3 3
1 2 4
7 6 5
3
1 2 1 2 2
1 2 1 3 4
2 3 2 3 4

6
5
1

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
	return x * f;
}

#define maxn 1010
#define maxm 300010
#define maxnode 9600010
#define bin 30
#define oo 2147483647

int n, m, X[maxn];

int ToT, rt[maxm], ch[maxnode][2], siz[maxnode];
void update(int& y, int x, int p, int val) {
	siz[y = ++ToT] = siz[x] + 1;
	if(p < 0) return ;
	memcpy(ch[y], ch[x], sizeof(ch[x]));
	update(ch[y][val>>p&1], ch[x][val>>p&1], p - 1, val);
	return ;
}
int lrt[maxn], rrt[maxn];
int qkth(int xl, int xr, int yl, int yr, int K) {
	for(int i = xl; i <= xr; i++) lrt[i] = rt[yl-1], rrt[i] = rt[yr];
	int ans = 0;
	for(int i = bin; i >= 0; i--) {
		int tmp = 0;
		for(int j = xl; j <= xr; j++)
			tmp += siz[ch[rrt[j]][X[j]>>i&1^1]] - siz[ch[lrt[j]][X[j]>>i&1^1]];
		if(K <= tmp) {
			for(int j = xl; j <= xr; j++)
				lrt[j] = ch[lrt[j]][X[j]>>i&1^1],
				rrt[j] = ch[rrt[j]][X[j]>>i&1^1];
			ans = ans << 1 | 1;
		}
		else {
			K -= tmp;
			for(int j = xl; j <= xr; j++)
				lrt[j] = ch[lrt[j]][X[j]>>i&1],
				rrt[j] = ch[rrt[j]][X[j]>>i&1];
			ans <<= 1;
		}
	}
	return ans;
}

int main() {
	n = read(); m = read();
	for(int i = 1; i <= n; i++) X[i] = read();
	for(int i = 1; i <= m; i++) update(rt[i], rt[i-1], bin, read());
	
	int q = read();
	while(q--) {
		int xl = read(), xr = read(), yl = read(), yr = read(), k = read();
		printf("%d\n", qkth(xl, xr, yl, yr, k));
	}
	
	return 0;
}