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 *a*<sub class="lower-index">*i*</sub> 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*.
The first line of input contains integer *n* (1 ≤ *n* ≤ 2 × 10<sup class="upper-index">5</sup>), the number of bears.

The second line contains *n* integers separated by space, *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>), heights of bears.
class Solution:
    def create_stack(self):
        stack=[]
        return stack

    def is_empty(self, stack):
        if stack==[]:
            return True
        return False

    def push(self, stack,new_ele):
        stack.append(new_ele)

    def pop(self, stack):
        if not self.is_empty(stack):
            x=stack.pop()
            return x
    
    def peek(self,stack):
        if not self.is_empty(stack):
            return stack[-1]
    
    def mikeFeet(self,heights,n):
        # we calculate pse and nses
        nse=[n-1]*n # init nses such that elems who don't have nse automatically get assigned the nse
        pse=[0]*n
        stack=self.create_stack()
        # figure out nses
        self.push(stack,0)
        for i in range(1,n):
            while not self.is_empty(stack) and heights[i]<heights[self.peek(stack)]:
                idx=self.pop(stack)
                nse[idx]=i-1
            self.push(stack,i)
        
        # empty stack for reuse
        while not self.is_empty(stack):
            self.pop(stack)
        
        # figure out pses
        self.push(stack,n-1)
        for i in range(n-2,-1,-1):
            while not self.is_empty(stack) and heights[i]<heights[self.peek(stack)]:
                idx=self.pop(stack)
                pse[idx]=i+1
            self.push(stack,i)
        
        sweep_lens=[nse[i]-pse[i]+1 for i in range(n)]

        ans=[0]*n # it is safe to use zero as marker as heights are >0
        for i in range(n):
            # 0-indexed
            ans[sweep_lens[i]-1]=max(ans[sweep_lens[i]-1],heights[i])

        for i in range(n-1,-1,-1):
            if ans[i]==0:
                ans[i]=ans[i+1]
        
        return ans

if __name__ == "__main__":
    import sys
    input=sys.stdin.readline
    n=int(input())
    heights=[int(i) for i in input().split()]

    soln=Solution()
    print(*soln.mikeFeet(heights,n))