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 <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <cassert>
#include <climits>
#include <string>
#include <bitset>
#include <cfloat>
#include <random>
#include <iomanip>
#include <unordered_set>
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<ll> vll;
typedef vector<pair<int,int> > vpii;
typedef vector<pair<ll,ll> > vpll;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<vc> vvc;
typedef vector<vs> vvs;
typedef vector<vll> vvll;
typedef map<int,int> mii;
typedef set<int> si;
#define rep(i,n) for(ll i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define arep(i,v) for(auto i : v)
#define irep(it, stl) for(auto it = stl.begin(); it != stl.end(); it++)
#define drep(i,n) for(int i = (n) - 1; i >= 0; --i)
#define fin(ans) cout << (ans) << '\n'
#define STLL(s) strtoll(s.c_str(), NULL, 10)
#define mp(p,q) make_pair(p, q)
#define pb(n) push_back(n)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define Sort(a) sort(a.begin(), a.end())
#define Rort(a) sort(a.rbegin(), a.rend())
#define MATHPI acos(-1)
#define fi first
#define se second
constexpr int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
constexpr int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
template <class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template <class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}
template <class T> inline void dump(T &v){irep(i, v){cout<<*i<<((i==--v.end())?'\n':' ');}}
inline string getline(){string s; getline(cin,s); return s;}
inline void yn(const bool b){b?fin("yes"):fin("no");}
inline void Yn(const bool b){b?fin("Yes"):fin("No");}
inline void YN(const bool b){b?fin("YES"):fin("NO");}
struct io{io(){ios::sync_with_stdio(false);cin.tie(0);}};
const int INF = INT_MAX;
constexpr ll LLINF = 1LL<<60;
constexpr ll MOD = 1000000007;
constexpr ld EPS = 1e-11;


int main() {
    
    int n;
    cin>>n;
    map<string,int> m;
    rep(i,n){
        int q;
        cin>>q;
        if(q==0){
            int x;
            string k;
            cin>>k>>x;
            m[k]=x;
        } else if(q==1) {
            string k;
            cin>>k;
            fin(m[k]);
        } else{
            string k;
            cin>>k;
            m.erase(k);
        }
    }
}