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
#include<bits/stdc++.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 1000000000000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;


map<string,int> MAP;

int main(){

	int num_query;
	scanf("%d",&num_query);

	int command,x;

	for(int loop = 0; loop < num_query; loop++){

		scanf("%d",&command);

		string tmp_str;
		string L,R;

		if(command == 0){

			cin >> tmp_str >> x;
			MAP[tmp_str] = x;

		}else if(command == 1){

			cin >> tmp_str;
			auto at = MAP.find(tmp_str);

			if(at == MAP.end()){

				printf("0\n");
			}else{

				printf("%d\n",MAP[tmp_str]);
			}

		}else if(command == 2){

			cin >> tmp_str;
			MAP.erase(tmp_str);

		}else{

			cin >> L >> R;

			auto at = MAP.lower_bound(L);
			auto right = MAP.upper_bound(R);

			while(at != right){

				printf("%s %d\n",at->first.c_str(),at->second);
				at++;
			}
		}
	}

	return 0;
}

