Codeforces Round #305 (Div. 2) D. Mike and Feet

D. Mike and Feet

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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 
学些数据结构才可以拯救自己吧，数据结构是省时的好东西哦，本来n^2可以变为nlgn甚至n，简直爆炸，单调栈就算能把复杂程度降到n的，太强辣，强行学习一波
其实我只是为了做出我们平台那道题，大佬讲让我用单调栈类似物，但是我并不了解这种数据结构啊，先mark下

#include <bits/stdc++.h>
using namespace std;
const int N=2e5+;
struct node
{
    int L, num;
}S;
stack <node> sta;
int a[N], ans[N];
int main ()
{
    int n;
    while (scanf ("%d", &n) != EOF)
    {
        for (int i=; i<n; i++)
            scanf ("%d", &a[i]);
 
        for (int i=; i<=n; i++)
        {
            int l = ;
 
            while (!sta.empty ())
            {
                S = sta.top ();
 
                if (S.num < a[i]) break;
 
                int lr = S.L + l;
                ans[lr] = max (ans[lr], S.num);
 
                l += S.L;
                sta.pop ();
            }
            sta.push((node){l+, a[i]});
        }
 
        for (int i=n; i>=; i--)
            ans[i] = max (ans[i], ans[i+]);
        for (int i=; i<n; i++)
            printf ("%d ", ans[i]);
        printf ("%d\n", ans[n]);
    }
    return ;
}