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 $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.

Constraints

* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by dump operations does not exceed $1,000,000$

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 $key$


or


3 $L$ $R$


where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.

Output

For each get operation, print the corresponding value.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys.

Example

Input

9
0 blue 4
0 red 1
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z


Output

1
4
0
0
white 5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define FOR(I,A,B) for(ll I = ll(A); I < ll(B); ++I)



int main(){
	int q,c,x;
	string key,L,R;
	cin >> q;
	map<string,int> ma;
	while(q--){
		cin >> c >> key;
		if(c==1){
			cout << ma[key] << "\n";
		}else if(c==0){
			cin >> x;
			ma[key] = x;
		}else if(c==2){
			//ma[key] = 0;
			ma.erase(key);
		}else if(c==3){
			L = key;
			cin >> R;
			auto it = ma.lower_bound(L);
			auto end = ma.upper_bound(R);
			for(;it!=end;it++){
				if((*it).second)cout << (*it).first << " " << (*it).second << "\n";
			}
		}
	}
}