D. Vitya and Strange Lesson

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 andmex([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, andm queries. Each query is characterized by one numberx and consists of the following consecutive steps:

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

Note that after each query the array changes.

Input

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

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

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

Output

For each query print the answer on a separate line.

Examples

Input

2 21 313

Output

10

Input

4 30 1 5 6124

Output

200

Input

5 40 1 5 6 71145

Output

2202

cf的d题，看题解都看了好久，主要是太弱了。

思路来自harlow_cheng 感谢

题意：给你一个数列，然后给你一个数字k，让你对数列中的每一个数字异或k，之后求出新数列中没出现过的最小自然数。

由于异或运算具有结合性，所以a^b^c^d=a^(b^c^d),所以你没有必要数组进行更新。

如果说给你一个值K，那么你对给定数列a中的元素进行异或k，将会得到一个新的数列A，同时我们定义b是a数列中没有出现过的自然数，这时我们再对b进行异或k，我们将得到一个B数列。B中的元素与A中的元素依旧存在如同b与a之间的关系，即B中的元素是A中未出现过的自然数。

明白了以上道理我们就可以把A全部插入01字典树，然后求一个异或k最小的数值就可以了。

AC代码如下，来自mengxiang000000 感谢

#include<stdio.h>#include<string.h>#include<map>#include<algorithm>#include<math.h>#include<stdlib.h>using namespace std;#define ll long long int#define maxn 2typedef struct tree{    tree *nex[maxn];    ll v;    ll val;} tree;tree root;void init(){    for(ll i=0; i<maxn; i++)    {        root.nex[i]=NULL;        root.v=0;        root.val=0;    }}void creat(char *str,int va){    int len=strlen(str);    tree *p=&root,*q;    for(int i=0; i<len; i++)    {        int id=str[i]-'0';        if(p->nex[id]==NULL)        {            q=(tree *)malloc(sizeof(root));            q->v=1;            for(int j=0; j<2; j++)            {                q->nex[j]=NULL;            }            p->nex[id]=q;        }        else        {            p->nex[id]->v++;        }        p=p->nex[id];        if(i==len-1)        {            p->val=va;        }    }}void find(char *str,ll query){    ll len=strlen(str);    tree *p=&root;    for(ll i=0; i<len; i++)    {        ll id=str[i]-'0';        if(p->nex[id]!=NULL)        {            p=p->nex[id];        }        else p=p->nex[1-id];        if(p==NULL)return ;    }    printf("%lld\n",p->val^query);}int main(){    ll n,q;    while(~scanf("%lld%lld",&n,&q))    {        init();        map<ll,ll>s;        for(ll i=1; i<=n; i++)        {            ll x;            scanf("%lld",&x);            if(s[x]==0)            {                s[x]=1;            }        }        for(ll i=0; i<=600000; i++)        {            if(s[i]==0)            {                char ss[25];                ll x=i;                 ss[21]='\0';                for(int j=20; j>=0; j--)                {                    if(x)                    {                        ss[j]=x%2+'0';                        x/=2;                    }                    else                    {                        ss[j]='0';                    }                }                creat(ss,i);            }        }        ll temp=0;        while(q--)        {            ll x;            scanf("%lld",&x);            temp^=x;            x=temp;            char ss[25];            ss[21]='\0';            for(int j=20; j>=0; j--)            {                if(x)                {                    ss[j]=x%2+'0';                    x/=2;                }                else                {                    ss[j]='0';                }            }            find(ss,temp);        }    }}