G - Vitya and Strange Lesson（数组-字典树）

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G - Vitya and Strange Lesson

CodeForces - 842D

Today at the lesson Vitya learned a very interesting function — mex. Mex of a sequence of numbers is the minimum non-negative number that is not present in the sequence as element. For example, mex([4, 33, 0, 1, 1, 5]) = 2 and mex([1, 2, 3]) = 0.

Vitya quickly understood all tasks of the teacher, but can you do the same?

You are given an array consisting of n non-negative integers, and m queries. Each query is characterized by one number x and consists of the following consecutive steps:

Perform the bitwise addition operation modulo 2 (xor) of each array element with the number x.
Find mex of the resulting array.

Note that after each query the array changes.

Input

First line contains two integer numbers n and m (1 ≤ n, m ≤ 3·105) — number of elements in array and number of queries.

Next line contains n integer numbers ai (0 ≤ ai ≤ 3·105) — elements of then array.

Each of next m lines contains query — one integer number x (0 ≤ x ≤ 3·105).

Output

For each query print the answer on a separate line.

Examples

Input

Output

1
0

Input

Output

2
0
0

Input

Output

题意 + 思路：

这两篇博客解释的非常好，我就不多说了

https://blog.csdn.net/brazy/article/details/77841433

https://blog.csdn.net/xzzf1024/article/details/79181290

解释：

主函数里再次求MEX的时候和上次的MEX值异或

num^=x; //与上次得到的值异或
printf("%d\n",num^Find_Trie(num));

这相当于再次输入的X与存在字典树的序列异或。（设存在字典树的序列a，则相当于a^x1^x2）;

个人理解+代码：

//MEX求数列中未出现过的最小自然数
//X异或任何一位结果不同
//与存在的数异或后 会再次成为这个N数组的数
//求MEX的值 会出现在 不在这N数组的数。
//再次求MEX的时候 和上次的MEX值异或 就可以了
//
#include <map>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int maxn=1e6;
int trie[5*maxn][2];
int vis[5*maxn];
int rt,tot;
void init()
{mem(vis,0);mem(trie,0);tot=0;
}
void Build_Trie(int y)
{rt=0;for(int i=31; i>=0; i--){int x=(y>>i)&1;if(!trie[rt][x])trie[rt][x]=++tot;rt=trie[rt][x];}vis[rt]=y;
}
int Find_Trie(int y)
{rt=0;int ans=0;for(int i=31; i>=0; i--){int x=(y>>i)&1;    //找到最小的异或值if(!trie[rt][x])rt=trie[rt][x^1];elsert=trie[rt][x];}return vis[rt];
}
map<int,int>mp;
int main()
{int n,m,y;while(~scanf("%d%d",&n,&m)){init();mp.clear();int x;for(int i=0; i<n; i++){scanf("%d",&x);mp[x]=1;}for(int i=0; i<=600005; i++){if(!mp[i])Build_Trie(i);}int num=0;for(int i=0; i<m; i++){scanf("%d",&x);num^=x;           //与上次得到的值异或printf("%d\n",num^Find_Trie(num));}}return 0;
}

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