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 <queue>
#include <vector>
#include <iostream>
#include <vector>
#include <cassert>
#include <algorithm>
#include <functional>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <deque>
#include <map>
using ll=long long;
using namespace std;
//const int INF = 7+(1e+9);
ll GCD(ll x, ll y)
{
    if (y == 0)
    {
        return x;
    }
    else
    {
        return GCD(y, x % y);
    }
}
ll LCM(ll x, ll y)
{
    return x / GCD(x, y) * y;
}
struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};
vector<vector<char>> vec(12,vector<char>(12));

int dx[]={0,-1,0,1};
int dy[]={1,0,-1,0};
int dfs(int y,int x){
    vec[y][x]='5';
    for(int i=0;i<4;i++){
        if(x+dx[i]<12&&x+dx[i]>=0&&y+dy[i]<12&&y+dy[i]>=0&&vec[dy[i]+y][dx[i]+x]=='1'){
            dfs(y+dy[i],x+dx[i]);
        }
    }
    return 0;
}
int main(){
    int n;cin>>n;
    map<string,int> mp;
    for(int i=0;i<n;i++){
        int x;cin>>x;
        if(x==0){
            string t;
            int y;
            cin>>t;
            cin>>y;
            mp[t]=y;
        }else if(x==1){
            string t;
            cin>>t;
            if(mp[t]){
                cout<<mp[t]<<endl;
            }else{
                cout<<0<<endl;
            }
        }else{
            string t;cin>>t;
            mp.erase(t);
        }
    }
}