package Others;

import java.util.Scanner;
import java.util.Arrays;

/**
 * To find triplet equals to given sum in complexity O(n*log(n))
 * 
 *
 * Array must be sorted 
 *
 * @author Ujjawal Joshi
 * @date 2020.05.18
 *
 * Test Cases:
 	Input:
 *	6 //Length of array
 	12 3 4 1 6 9
 	target=24
 *	Output:3 9 12
 *	Explanation: There is a triplet (12, 3 and 9) present
	in the array whose sum is 24. 
 *
 *

 */


class threesum{
	public static void main(String args[])
	{
		Scanner sc =new Scanner(System.in);
		int n=sc.nextInt(); //Length of an array

		int a[]=new int[n];

		for(int i=0;i<n;i++)
		{
			a[i]=sc.nextInt();
		}
		System.out.println("Target");
		int n_find=sc.nextInt();

		Arrays.sort(a);	// Sort the array if array is not sorted

		for(int i=0;i<n;i++){	

			int l=i+1,r=n-1;

			while(l<r){
				if(a[i]+a[l]+a[r]==n_find) {System.out.println(a[i]+" "+ a[l]+" "+a[r]);break;} //if you want all the triplets write l++;r--; insted of break; 
				else if(a[i]+a[l]+a[r]<n_find) l++;
				else r--;
			}
		}

		sc.close();
		
	}
}