Codeforces Round #573 Tokitsukaze, CSL and Stone Game

B. Tokitsukaze, CSL and Stone Game

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Tokitsukaze and CSL are playing a little game of stones.

In the beginning, there are n piles of stones, the i-th pile of which has ai stones. The two players take turns making moves. Tokitsukaze moves first. On each turn the player chooses a nonempty pile and removes exactly one stone from the pile. A player loses if all of the piles are empty before his turn, or if after removing the stone, two piles (possibly empty) contain the same number of stones. Supposing that both players play optimally, who will win the game?

Consider an example: n=3 and sizes of piles are a1=2, a2=3, a3=0. It is impossible to choose the empty pile, so Tokitsukaze has two choices: the first and the second piles. If she chooses the first pile then the state will be [1,3,0] and it is a good move. But if she chooses the second pile then the state will be [2,2,0] and she immediately loses. So the only good move for her is to choose the first pile.

Supposing that both players always take their best moves and never make mistakes, who will win the game?

Note that even if there are two piles with the same number of stones at the beginning, Tokitsukaze may still be able to make a valid first move. It is only necessary that there are no two piles with the same number of stones after she moves.

Input

The first line contains a single integer n (1≤n≤105) — the number of piles.

The second line contains n integers a1,a2,…,an (0≤a1,a2,…,an≤109), which mean the i-th pile has ai stones.

Output

Print “sjfnb” (without quotes) if Tokitsukaze will win, or “cslnb” (without quotes) if CSL will win. Note the output characters are case-sensitive.

Examples

input

0
output
cslnb

input

2
1 0

output

cslnb

input

2
2 2

output

sjfnb

input

3
2 3 1

output

sjfnb

Note

In the first example, Tokitsukaze cannot take any stone, so CSL will win.
In the second example, Tokitsukaze can only take a stone from the first pile, and then, even though they have no stone, these two piles will have the same number of stones, which implies CSL will win.
In the third example, Tokitsukaze will win. Here is one of the optimal ways:
Firstly, Tokitsukaze can choose the first pile and take a stone from that pile.
Then, CSL can only choose the first pile, because if he chooses the second pile, he will lose immediately.
Finally, Tokitsukaze can choose the second pile, and then CSL will have no choice but to lose.
In the fourth example, they only have one good choice at any time, so Tokitsukaze can make the game lasting as long as possible and finally win.

This title is a game of chance, there are n heap of stone, the two successive operations, each can be removed in a pile, and after the operation if the same amount of time a two heap, then lose. Piles of the same, if it can not have the same after an operation, the game can proceed when there is initial.
Start-win situation first appeared there:

• Three identical
• Two pairs and more of the same
• Two 0
• A pair of identical, but the same, the smaller number than it is present 1
• Is less than the sum n*(n-1)/2, this occurs will appear above three cases.

Not the same as the above without the occurrence of the stack in one operation, thus losing the initial state.
In addition to losing unexpected state, the operation will be the final state 0 1 2 3....n-1which is not the last form of operation, any operation will appear two identical stacks. Because each only one of the pair of operation, but before the final state, always find operation. Such operational quantity is sum - n*(n-1)/2then distributed to the parity out who is the winner.
The following is the code of java AC

import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;
 
public class Main {
	static StreamTokenizer st = new StreamTokenizer(new BufferedInputStream(System.in));
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter pr = new PrintWriter(new BufferedOutputStream(System.out));
	static Scanner sc = new Scanner(System.in);
 
	public static void main(String[] args) throws NumberFormatException, IOException {
		long sum = 0;
		long n = sc.nextLong();
		HashSet<Long> hs = new HashSet<>();
		ArrayList<Long> al = new ArrayList<>();
		for (int i = 0; i < n; i++) {
			long te = sc.nextLong();
			sum += te;
			if(!hs.add(te)) {
				al.add(te);
			}
		}
		for (Long ye : al) {
			if(hs.contains(ye-1)||ye==0) {
				System.out.println("cslnb");
				System.exit(0);
			}
		}
		if (hs.size() < n - 1) {
			System.out.println("cslnb");
		} else {
			sum-=n*(n-1)/2;
			if (sum < 0) {
				System.out.println("cslnb");
			} else if ((sum & 1) == 0) {
				System.out.println("cslnb");
			} else {
				System.out.println("sjfnb");
			}
		}
	}
 
	private static int nextInt() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return (int) st.nval;
	}
}