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 $M$. If there is an element with $key$, replace the corresponding value with $x$.
* get($key$): Print the value 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 letter
* For a get operation, the element with the specified key exists in $M$.

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 $key$


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

Output

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

Example

Input

7
0 blue 4
0 red 1
0 white 5
1 red
1 blue
0 black 8
1 black


Output

1
4
8

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <string>
#include <fstream>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <string>
#include <set>
#include <map>
using namespace std;
typedef long long ll;

// #define DEBUG

void printarray(int n, int a[])
{
    for (int i = 1; i < n + 1; i++)
    {
        cout << a[i];
        if (i != n)
            cout << " ";
    }
    cout << endl;
}

int cycleanalysis(int n, int a[])
{
    int pos, ret, next;
    pos = 0, ret = 0;
    for (int i = 1; i < n + 1; i++)
    {
        pos = i;
        while (a[pos] != pos)
        {
            ret += (a[pos] + a[a[pos]]);
            next = a[pos];
            swap(a[pos], a[a[pos]]);
            pos = next;
        }
    }

    return ret;
}

int main()
{

#ifdef DEBUG
    cout << "DEBUG MODE" << endl;
    ifstream in("input.txt"); //for debug
    cin.rdbuf(in.rdbuf());    //for debug
#endif

    int q, op, x;
    string key;
    cin >> q;
    map<string, int> mp;
    for (int i = 0; i < q; i++)
    {
        cin >> op >> key;
        if (op == 0)
        {
            cin >> x;
            if (mp.count(key))
                mp[key] = x;
            else
                mp.insert(make_pair(key, x));
        }
        else if (op == 1)
        {
            cout << (mp.count(key) ? mp.at(key) : 0) << endl;
        }
        else if (op == 2)
        {
            mp.erase(key);
        }
    }

    return 0;
}

// getline(cin,s);
// a = strtok(s, ",");  ","で分解
// for (int i = 0; i < n; i++)