Problem: For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique.

* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print the value with the specified $key$. Print 0 if there is no such element.
* delete($key$): Delete the element with the specified $key$.

Constraints

* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consits of lower case letters

Input

The input is given in the following format.


$q$
$query_1$
$query_2$
:
$query_q$


Each query $query_i$ is given by


0 $key$ $x$


or


1 $key$


or


2 $key$


where the first digits 0, 1 and 2 represent insert, get and delete operations respectively.

Output

For each get operation, print an integer in a line.

Example

Input

8
0 blue 4
0 red 1
0 white 5
1 red
1 blue
2 red
1 black
1 red


Output

1
4
0
0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;
#define rep_i(i, n) for (int i = 0; i < (n); ++i)
#define rep_ll(i, n) for (long long i = 0; i < (n); ++i)
#define r_rep_i(i, start, end) for (int i = (start); i < (end); ++i)
#define r_rep_ll(i, start, end) for (long long i = (start); i < (end); ++i)
#define debug_vi(v) copy((v).begin(), (v).end(), ostream_iterator<int>(cout, " "));
#define debug_vll(v) copy((v).begin(), (v).end(), ostream_iterator<long long>(cout, " "));
#define debug_vd(v) copy((v).begin(), (v).end(), ostream_iterator<double>(cout, " "));
#define sort_v(v) sort((v).begin(), (v).end()); //昇順
#define d_sort_vi(v) sort((v).begin(), (v).end(), greater<int>()); //降
#define d_sort_vd(v) sort((v).begin(), (v).end(), greater<double>()); //昇順
#define say(t) cout << (t) << endl;
#define sum_vi(v) accumulate((v).begin(), (v).end(), 0);
#define sum_vll(v)accumulate((v).begin(), (v).end(), 0LL);

int main(){
  map<string, int> m;
  int q; cin >> q;
  int command;
  string key;
  int x;
  rep_i(i, q){
    cin >> command;
    if(command == 0){
      cin >> key >> x;
      auto it = m.find(key);
      if(it == m.end()){
        m.insert(make_pair(key, x));
      }else{
        m[key] = x;
      }
    }else if(command == 1){
      cin >> key;
      auto it = m.find(key);
      if(it == m.end()){
        say(0);
      }else{
        say(m.at(key));
      }
    }else if(command == 2){
      cin >> key;
      auto it = m.find(key);
      if(it != m.end()){
        m.erase(key);
      }
    }
  }
}