#include <bits/stdc++.h>
using namespace std;
int q;
string s[200001];
int x[200001];
string bits[200001];
int uk = 0;
int next1[10000001][2];
int prev1[10000001];
int cnt[10000001];
int root = 0;
map<string, int> where;
string getbits(int x) {
  string s;
  for (int i = 0; i < 32; i++) {
    if (x & 1) {
      s += "1";
    } else {
      s += "0";
    }
    x >>= 1;
  }
  reverse(s.begin(), s.end());
  return s;
}
void add1(string s, int t) {
  int cur = root;
  for (int i = 0; i < (int)s.size(); i++) {
    if (next1[cur][s[i] - '0']) {
      cur = next1[cur][s[i] - '0'];
      cnt[cur] += t;
    } else {
      uk++;
      next1[cur][s[i] - '0'] = uk;
      prev1[uk] = cur;
      cur = uk;
      cnt[cur] += t;
    }
  }
  where[s] = cur;
}
void delete1(string s) {
  int cur = where[s];
  while (cur) {
    cnt[cur]--;
    cur = prev1[cur];
  }
}
int getnum(string a) {
  int ans = 0;
  for (int i = 0; i < 32; i++) {
    ans *= 2;
    ans += a[i] - '0';
  }
  return ans;
}
int findans(string s) {
  int cur = root;
  string ans;
  for (int i = 0; i < (int)s.size(); i++) {
    if (next1[cur][1 - (s[i] - '0')] && cnt[next1[cur][1 - (s[i] - '0')]]) {
      cur = next1[cur][1 - (s[i] - '0')];
      ans += '1';
    } else {
      cur = next1[cur][s[i] - '0'];
      ans += '0';
    }
  }
  return getnum(ans);
}
int main() {
  cin >> q;
  add1(getbits(0), 1);
  for (int i = 0; i < q; i++) {
    cin >> s[i];
    scanf("%d", &x[i]);
    bits[i] = getbits(x[i]);
    add1(bits[i], 0);
  }
  for (int i = 0; i < q; i++) {
    if (s[i] == "+") {
      add1(bits[i], 1);
    } else {
      if (s[i] == "-") {
        delete1(bits[i]);
      } else {
        printf("%d\n", findans(bits[i]));
      }
    }
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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 <image>, 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 xi (1 ≤ xi ≤ 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 xi 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

Note

After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.

The answer for the sixth query is integer <image> — maximum among integers <image>, <image>, <image>, <image> and <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int q;
string s[200001];
int x[200001];
string bits[200001];
int uk = 0;
int next1[10000001][2];
int prev1[10000001];
int cnt[10000001];
int root = 0;
map<string, int> where;
string getbits(int x) {
  string s;
  for (int i = 0; i < 32; i++) {
    if (x & 1) {
      s += "1";
    } else {
      s += "0";
    }
    x >>= 1;
  }
  reverse(s.begin(), s.end());
  return s;
}
void add1(string s, int t) {
  int cur = root;
  for (int i = 0; i < (int)s.size(); i++) {
    if (next1[cur][s[i] - '0']) {
      cur = next1[cur][s[i] - '0'];
      cnt[cur] += t;
    } else {
      uk++;
      next1[cur][s[i] - '0'] = uk;
      prev1[uk] = cur;
      cur = uk;
      cnt[cur] += t;
    }
  }
  where[s] = cur;
}
void delete1(string s) {
  int cur = where[s];
  while (cur) {
    cnt[cur]--;
    cur = prev1[cur];
  }
}
int getnum(string a) {
  int ans = 0;
  for (int i = 0; i < 32; i++) {
    ans *= 2;
    ans += a[i] - '0';
  }
  return ans;
}
int findans(string s) {
  int cur = root;
  string ans;
  for (int i = 0; i < (int)s.size(); i++) {
    if (next1[cur][1 - (s[i] - '0')] && cnt[next1[cur][1 - (s[i] - '0')]]) {
      cur = next1[cur][1 - (s[i] - '0')];
      ans += '1';
    } else {
      cur = next1[cur][s[i] - '0'];
      ans += '0';
    }
  }
  return getnum(ans);
}
int main() {
  cin >> q;
  add1(getbits(0), 1);
  for (int i = 0; i < q; i++) {
    cin >> s[i];
    scanf("%d", &x[i]);
    bits[i] = getbits(x[i]);
    add1(bits[i], 0);
  }
  for (int i = 0; i < q; i++) {
    if (s[i] == "+") {
      add1(bits[i], 1);
    } else {
      if (s[i] == "-") {
        delete1(bits[i]);
      } else {
        printf("%d\n", findans(bits[i]));
      }
    }
  }
  return 0;
}

```