#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const long long INF = 2e18;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int lcm(int a, int b) { return (a * b) / gcd(a, b); }
void debug() { cout << "DEBUG" << endl; }
int exp(int a, int b) {
  int result = 1;
  while (b > 0) {
    if (b & 1) result = result * a;
    b >>= 1;
    a = a * a;
  }
  return result;
}
int exp(int a, int b, int m) {
  int result = 1;
  while (b > 0) {
    if (b & 1) result = result * a % m;
    b >>= 1;
    a = a * a % m;
  }
  return result % m;
}
struct trie {
  struct node {
    vector<int> children;
    bool end;
    node() {
      children.assign(2, -1);
      end = false;
    }
  };
  vector<node> root;
  trie() {
    node x;
    root.push_back(x);
  }
  void insert(int x) {
    int v = 0;
    for (int i = 30; i >= 0; i--) {
      int bit = x >> i & 1;
      if (root[v].children[bit] == -1) {
        root[v].children[bit] = root.size();
        root.push_back(node());
      }
      v = root[v].children[bit];
    }
    root[v].end = true;
  }
  int calc(int x) {
    int v = 0;
    int ans = 0;
    for (int i = 30; i >= 0; i--) {
      int bit = !(x >> i & 1);
      if (root[v].children[bit] == -1) {
        bit = !bit;
        v = root[v].children[bit];
      } else {
        ans += 1 << i;
        v = root[v].children[bit];
      }
    }
    return ans;
  }
  int cnt(node &x) {
    int ans = 0;
    for (int &i : x.children) {
      ans += (i != -1);
    }
    return ans;
  }
  bool del(int x, int index = 30, int v = 0) {
    if (index == -1) {
      if (!root[v].end) return false;
      root[v].end = false;
      return cnt(root[v]) == 0;
    }
    int bit = x >> index & 1;
    if (root[v].children[bit] == -1) return false;
    bool ans = del(x, index - 1, root[v].children[bit]);
    if (ans) {
      root[v].children[bit] = -1;
      return cnt(root[v]) == 0;
    }
    return false;
  }
  void consertar() {
    while (cnt(root.back()) == 0 and !root.back().end) {
      root.pop_back();
    }
  }
};
trie tr;
void solve() {
  tr.insert(0ll);
  int q;
  cin >> q;
  map<int, int> m;
  m[0] = 1;
  while (q--) {
    char c;
    cin >> c;
    int x;
    cin >> x;
    if (c == '+') {
      if (m[x] == 0) tr.insert(x);
      tr.consertar();
      m[x]++;
    } else if (c == '-') {
      m[x]--;
      if (m[x] == 0) tr.del(x);
    } else {
      cout << tr.calc(x) << endl;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Your challenge is to write 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;
const int MOD = 1e9 + 7;
const long long INF = 2e18;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int lcm(int a, int b) { return (a * b) / gcd(a, b); }
void debug() { cout << "DEBUG" << endl; }
int exp(int a, int b) {
  int result = 1;
  while (b > 0) {
    if (b & 1) result = result * a;
    b >>= 1;
    a = a * a;
  }
  return result;
}
int exp(int a, int b, int m) {
  int result = 1;
  while (b > 0) {
    if (b & 1) result = result * a % m;
    b >>= 1;
    a = a * a % m;
  }
  return result % m;
}
struct trie {
  struct node {
    vector<int> children;
    bool end;
    node() {
      children.assign(2, -1);
      end = false;
    }
  };
  vector<node> root;
  trie() {
    node x;
    root.push_back(x);
  }
  void insert(int x) {
    int v = 0;
    for (int i = 30; i >= 0; i--) {
      int bit = x >> i & 1;
      if (root[v].children[bit] == -1) {
        root[v].children[bit] = root.size();
        root.push_back(node());
      }
      v = root[v].children[bit];
    }
    root[v].end = true;
  }
  int calc(int x) {
    int v = 0;
    int ans = 0;
    for (int i = 30; i >= 0; i--) {
      int bit = !(x >> i & 1);
      if (root[v].children[bit] == -1) {
        bit = !bit;
        v = root[v].children[bit];
      } else {
        ans += 1 << i;
        v = root[v].children[bit];
      }
    }
    return ans;
  }
  int cnt(node &x) {
    int ans = 0;
    for (int &i : x.children) {
      ans += (i != -1);
    }
    return ans;
  }
  bool del(int x, int index = 30, int v = 0) {
    if (index == -1) {
      if (!root[v].end) return false;
      root[v].end = false;
      return cnt(root[v]) == 0;
    }
    int bit = x >> index & 1;
    if (root[v].children[bit] == -1) return false;
    bool ans = del(x, index - 1, root[v].children[bit]);
    if (ans) {
      root[v].children[bit] = -1;
      return cnt(root[v]) == 0;
    }
    return false;
  }
  void consertar() {
    while (cnt(root.back()) == 0 and !root.back().end) {
      root.pop_back();
    }
  }
};
trie tr;
void solve() {
  tr.insert(0ll);
  int q;
  cin >> q;
  map<int, int> m;
  m[0] = 1;
  while (q--) {
    char c;
    cin >> c;
    int x;
    cin >> x;
    if (c == '+') {
      if (m[x] == 0) tr.insert(x);
      tr.consertar();
      m[x]++;
    } else if (c == '-') {
      m[x]--;
      if (m[x] == 0) tr.del(x);
    } else {
      cout << tr.calc(x) << endl;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    solve();
  }
}

```