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<algorithm>
#include<climits>
#include<cmath>
#include<cstring>
#include<deque>
#include<functional>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<utility>
#include<vector>

using namespace std;

using lint = long long;
using P = pair<int, int>;
using LLP = pair<long long, long long>;

#define REP(i, x, n) for(int i = (x), i##_len = int(n) ; i < i##_len ; ++i)
#define rep(i, n) for(int i = 0, i##_len = int(n) ; i < i##_len ; ++i)
#define reps(i, n) for(int i = 1, i##_len = int(n) ; i <= i##_len ; ++i)
#define rrep(i, n) for(int i = int(n) - 1 ; i >= 0 ; --i)
#define rreps(i, n) for(int i = int(n) ; i > 0 ; --i)
#define SORT(x) sort((x).begin(), (x).end())
#define SORT_INV(x) sort((x).rbegin(), (x).rend())
#define TWINS(x) cout << ((x) ? "Yay!" : ":(") << endl

const int IINF = (1 << 30) - 1;
const long long LLINF = 1LL << 61;
const int dx4[] = {1, 0, -1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy8[] = {0, -1, -1, -1, 0, 1, 1, 1};
const double EPS = 1e-8;

template<typename T>
bool chmax(T &a, T b){
    if(a < b){
        a = b;
        return true;
    }
    return false;
}

template<typename T>
bool chmin(T &a, T b){
    if(b < a){
        a = b;
        return true;
    }
    return false;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int q;
    cin >> q;

    map<string, int> mp;
    while(q--){
        int m, x;
        string k;
        cin >> m;

        if(m == 0){
            cin >> k >> x;
            mp[k] = x;
        }else if(m == 1){
            cin >> k;
            cout << mp[k] << endl;
        }else if(m == 2){
            cin >> k;
            auto it = mp.find(k);
            if(it != mp.end()){
                mp.erase(it);
            }
        }
    }

    return 0;
}