Problem:
Score : 300 points

Problem Statement

You are given a sequence of length N: A_1, A_2, ..., A_N.
For each integer i between 1 and N (inclusive), answer the following question:

Find the maximum value among the N-1 elements other than A_i in the sequence.

Constraints

2 \leq N \leq 200000

1 \leq A_i \leq 200000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print N lines. The i-th line (1 \leq i \leq N) should contain the maximum value among the N-1 elements other than A_i in the sequence.

Sample Input 1

3
1
4
3

Sample Output 1

4
3
4

The maximum value among the two elements other than A_1, that is, A_2 = 4 and A_3 = 3, is 4.

The maximum value among the two elements other than A_2, that is, A_1 = 1 and A_3 = 3, is 3.

The maximum value among the two elements other than A_3, that is, A_1 = 1 and A_2 = 4, is 4.

Sample Input 2

2
5
5

Sample Output 2

5
5

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        List<Integer> list = new ArrayList<Integer>();
        List<Integer> lo = new ArrayList<Integer>();
        
        for (int i = 0; i < N; i++) {
            int t = sc.nextInt();
            list.add( t);
            lo.add( t);
        }
        Collections.sort(list);
        int max = list.get(N-1);
        int smax = -1;
        boolean fl_dmax = false;
        if(list.get(N-2) == max)    fl_dmax = true;
        for(int i=list.size()-1; i>=0; i--) {
            if(list.get(i) < max) {
                smax = list.get(i);
                break;
            }
        }

        for (int i = 0; i < lo.size(); i++) {
            if( lo.get(i) != max ) {
                System.out.println(max);
            }else{
                System.out.println( fl_dmax? max: smax );
            }
        }
    }
}