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

[b]D. Mike and Feet[/b]

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

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=2e5+10,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=1e9+7;
///   数组大小
int a
,ans
;
int l
,r
,s
;
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
int st=0;
s[++st]=0;
for(int i=1;i<=n;i++)
{
while(a[s[st]]>=a[i])st--;
l[i]=s[st];
s[++st]=i;
}
st=0;
s[++st]=n+1;
for(int i=n;i>=1;i--)
{
while(a[s[st]]>=a[i])st--;
r[i]=s[st];
s[++st]=i;
}
for(int i=1;i<=n;i++)
{
int len=r[i]-l[i]-1;
ans[len]=max(ans[len],a[i]);
}
for(int i=n;i>=1;i--)
ans[i]=max(ans[i],ans[i+1]);
for(int i=1;i<=n;i++)
printf("%d ",ans[i]);
return 0;
}