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 JAVA):

Solution:
import java.util.Scanner;
import java.util.SortedMap;
import java.util.TreeMap;

class Main {

	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)){
			int times = sc.nextInt();
			TreeMap<String, Long> map = new TreeMap<>();
			for (int i = 0; i<times; i++){
				int query = sc.nextInt();
				String key = sc.next();
				switch (query) {
				case 0:
					long x = sc.nextLong();
					map.put(key,x);
					break;
				case 1:
					if (map.get(key)!=null) {
						System.out.println(map.get(key));
					} else System.out.println(0);
					break;
				case 2:
					map.remove(key);
					break;
				case 3:
					String large = sc.next();
					SortedMap<String,Long> submap = map.subMap(key,true,large,true);
					for (String s : submap.keySet()) {
						System.out.println(s + " " + submap.get(s));   //回数があるだけ出力
					}
					break;
				}
			}
		}
	}
}