Codeforces Round #430 (Div. 2) D. Vitya and Strange Lesson

地址：http://codeforces.com/contest/842/problem/D

题目：

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

思路：

　　很明显的trie树题，本来以为写完就ac了的，没想到一直wa6.。。。

　　今天起来后发现是把重复数字插入了，应该hash去重的。

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 int hs[1000000];
 5 struct Trie
 6 {
 7     int root, tot, next[1000005][2], cnt[1000005], end[1000005];
 8 
 9     inline int Newnode()
10     {
11         memset(next[tot], -1, sizeof(next[tot]));
12         cnt[tot] = 0;
13         end[tot] = 0;
14         return tot ++;
15     }
16 
17     inline void Init()
18     {
19         tot = 0;
20         root = Newnode();
21     }
22 
23     inline int Insert(int x)
24     {
25         int p = root;
26         cnt[p] ++;
27         for(int i = 31; i >= 0; i --)
28         {
29             int idx = ((1 << i) & x) ? 1 : 0;
30             if(next[p][idx] == -1)
31                 next[p][idx] = Newnode();
32             p = next[p][idx];
33             cnt[p] ++;
34         }
35         end[p] = x;
36         return 1;
37     }
38     inline int Search(int x)
39     {
40         int p = root;
41         int sum[50],ret=0;
42         sum[0]=1;
43         for(int i=1;i<31;i++)
44             sum[i]=sum[i-1]*2;
45         for(int i = 31; i >= 0; i --)
46         {
47             int idx = ((1 << i) & x) ? 1 : 0;
48             if(idx == 0)
49             {
50                 if(next[p][0]==-1) return ret;
51                 else if(cnt[next[p][0]]==sum[i])
52                     p = next[p][1],ret+=sum[i];
53                 else
54                     p = next[p][0];
55             }
56             else
57             {
58                 if(next[p][1]==-1) return ret;
59                 else if(cnt[next[p][1]]==sum[i])
60                     p = next[p][0],ret+=sum[i];
61                 else
62                     p = next[p][1];
63             }
64             if(p==-1) return ret;
65         }
66         return ret;
67     }
68 }tr;
69 
70 int main(void)
71 {
72     int n,m;
73     cin>>n>>m;
74     tr.Init();
75     for(int i=1,x;i<=n;i++)
76         scanf("%d",&x),hs[x]?0:hs[x]=tr.Insert(x);
77     for(int i=1,x,ls=0;i<=m;i++)
78         scanf("%d",&x),printf("%d
",tr.Search(ls=(x^ls)));
79     return 0;
80 }