字典树：又称单词查找树，Trie树，是一种树形结构，是一种哈希树的变种。典型应用是用于统计，排序和保存大量的字符串（但不仅限于字符串，所以经常被搜索引擎系统用于文本词频统计。他的优点是：利用字符串的公共前缀来减少查询时间，最大限度地减少无畏的字符串比较，查询效率比哈希高）//具体见百度百科

基本操作：查找，插入和删除。

例题：

D. Vasiliy's Multiset

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Author has gone out of the stories about Vasiliy, so here is just a formal task description.

You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:

1. "+ x" — add integer x to multiset A.
2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
3. "? x" — you are given integer x and need to compute the value                          , i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.

Multiset is a set, where equal elements are allowed.

Input

The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.

Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer x_i (1 ≤ x_i ≤ 10⁹). It's guaranteed that there is at least one query of the third type.

Note, that the integer 0 will always be present in the set A.

Output

For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer x_i and some integer from the multiset A.

Example

input

10

+ 8

+ 9

+ 11

+ 6

+ 1

? 3

- 8

? 3

? 8

? 11

output

11

10

14

13

 

 

就是求^的最大值。01字典树解决。

//搞不懂N是怎样算的 我让N=（10e5）mle了

AC代码：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;typedef long long LL;const int N = (2e5)*30;const int INF = 0x3f3f3f3f;int ch[N][2],cnt[N],tot,n;int newnode(){    ++tot;    memset(ch[tot],-1,sizeof(ch[tot]));    return tot;}void add(int x,int t){    int now=0;    for(int i=29; i>=0; --i)    {        int nx=(x&(1<
         
          =0; --i)    {        int nx=(x&(1<