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 <iostream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <string>
#include <functional>
#include <map>

using namespace std;

int main()
{
    int q;
    cin >> q;

    map<string, int> S;
    int cmd, x;
    string key;
    while (q--)
    {
        cin >> cmd;
        if (cmd == 0)
        {
            cin >> key >> x;
            S[key] = x;
            // cout << S.size() << endl;
        }
        else if (cmd == 1)
        {
            cin >> key;
            // cout << (S.find(x) != S.end()) << endl;

            if (S.find(key) != S.end())
                cout << S[key] << endl;
            else
                cout << 0 << endl;

            //由于不存在键时，会以默认值插入,之后则无法插入相同键，所以不可用
            //  cout << S[key] << endl;
        }
        else if (cmd == 2)
        {
            cin >> key;
            S.erase(key);
            // remove(S.begin(), S.end(), x);
        }
        else
        {
            string L, R;
            cin >> L >> R;

            map<string ,int>::iterator it1 = S.lower_bound(L), it2 = S.upper_bound(R);
            while (it1 != it2)
            {
                cout << it1->first << ' ' << it1->second << endl;
                it1++;
            }

            //法2， 超时
            // map<string, int>::iterator it = S.begin();
            // while (it != S.end())
            // {
            //     if (it->first >= L && it->first <= R)
            //     cout << it->first << ' ' << it->second << endl;
            //     it++;                
            // }

        }

    }
}