Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high. 

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A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.

Mike is a curious to know for each x such that 1 ≤ x ≤ n the maximum strength among all groups of size x.

Input

The first line of input contains integer n (1 ≤ n ≤ 2 × 105), the number of bears.

The second line contains n integers separated by space, a1, a2, ..., an (1 ≤ ai ≤ 109), heights of bears.

Output

Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.

Examples

Input

10
1 2 3 4 5 4 3 2 1 6


Output

6 4 4 3 3 2 2 1 1 1 
//package Codeforces;

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

/**
 * Created by rajanW(683281) on 28-May-15.
 * at 11:18 AM
 */
public class Main{

    static int n;
    static int[] a;

    //failed time limit in testcase 7
    static void solveBruteForce(){
        int[] ans = new int[n+1];
        StringBuilder sb = new StringBuilder();
        for(int x=1;x<=n;x++){
            int finMax = 0;
            for(int i=0;i+x<=n;i++){//start of len x
                int j = i+x;int temp = a[i];
                for(int k=i;k<j;k++){
                    temp=Math.min(temp, a[k]);
                }
                finMax=Math.max(temp,finMax);
            }
            ans[x]=finMax;
            sb.append(ans[x]+" ");
        }
        System.out.println(sb.toString().trim());
    }


    static void solveLinearTime(){
        StringBuilder sb = new StringBuilder();
        int[] Ls = new int[n+2];//L[i] = smallest j such that aj<ai
        Stack<Integer> st = new Stack<Integer>();
        for(int i=1;i<=n;i++){
            while(!st.isEmpty()&&a[st.peek()]>=a[i]){st.pop();}
            if(st.isEmpty()){Ls[i]=0;}
            else{Ls[i]=st.peek();}
            st.push(i);
        }
        int[] Rs = new int[n+2];//R[i]: largest j such that aj<ai
        Stack<Integer> ss = new Stack<Integer>();
        ss.push(n+1);
        for(int i=n;i>0;i--){
            while(!ss.isEmpty()&&a[ss.peek()]>=a[i]){ss.pop();}
            if(ss.isEmpty()){Rs[i]=0;}
            else{Rs[i]=ss.peek();}
            ss.push(i);
        }

        // a[i] can be minimum element in array of size 1,2,..,r-l-1

//        System.out.println(Arrays.toString(Ls));
//        System.out.println(Arrays.toString(Rs));

        int[] ans = new int[n+1];
        for(int i=1;i<=n;i++){
            int len = Rs[i]-Ls[i]-1;// what can be max strength of this sized group ? a[i] or what already had
            ans[len] = Math.max(ans[len],a[i]);
        }
//        System.out.println(Arrays.toString(ans));

        for(int i=n-1;i>0;i--){ //for those size for which there was no 'len' value
            ans[i]=Math.max(ans[i],ans[i+1]);
        }
        for(int i=1;i<=n;i++){sb.append(ans[i]+" ");}
        System.out.println(sb.toString().trim());
    }



    public static void main(String args[]) throws Exception{
        Scanner sc = new Scanner(System.in);
        n = Integer.parseInt(sc.nextLine().trim());
        a = new int[n+2];
        String[] sa = sc.nextLine().split(" ");
        for(int i=1;i<=n;i++){
            a[i]=Integer.parseInt(sa[i-1]);
        }
        //solveBruteForce();
        solveLinearTime();
    }

}
