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 <string>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <stack>
#include <numeric>
#include <algorithm>
#include <stdio.h>
#include <bitset>
#include <limits.h>
#include <complex>
#include <deque>
#include <iomanip>
using namespace std;
#define MM 1000000000
#define MOD MM+7
#define pi pair<int32,int32>
#define pl pair<int64,int64>
#define chmax(a,b) (a<b ? a=b:0)
#define chmin(a,b) (a>b ? a=b:0)
#define en cout << endl //セミコロンつけろ
const long double PI = acos(-1);
const long long INF = 1e15;
int dx[8] = {-1,0,1,0,-1,-1,1,1};
int dy[8] = {0,-1,0,1,-1,1,1,-1};
template<typename T>
T GCD(T u,T v){
    return v ? GCD(v,u%v) : u;
}
template<typename T>
T LCM(T x,T y){
    T gc = GCD(x,y);
    return x*y/gc;
}

long long pow_mod(long long n,long long k,long long m){
    if (k == 0){
        return 1;
    } else if (k % 2 == 1){
        return pow_mod(n,k-1,m)*n % m;
    } else {
        long long t = pow_mod(n,k/2,m);
        return t*t % m;
    }
}


int main(){
    int q; cin >> q;
    map<string,int> mp;
    for(int i = 0; i < q; i++){
        int n; string s; cin >> n >> s;
        if (n == 0){
            int k; cin >> k;
            mp[s] = k;
        } else if (n == 1){
            if (mp[s] > 0) cout << mp[s] << endl;
            else cout << 0 << endl;
        } else {
            mp[s] = 0;
        }
    }
}