【題目描述】

Rick and Morty want to find MR. PBH and they can't do it alone. So they need of Mr. Meeseeks. They Have generated n Mr. Meeseeks, standing in a line numbered from 1 to n. Each of them has his own color. i-th Mr. Meeseeks' color is ai.Rick and Morty are gathering their army and they want to divide Mr. Meeseeks into some squads. They don't want their squads to be too colorful, so each squad should have Mr. Meeseeks of at most k different colors. Also each squad should be a continuous subarray of Mr. Meeseeks in the line. Meaning that for each 1?≤?i?≤?e?≤?j?≤?n, if Mr. Meeseeks number i and Mr. Meeseeks number j are in the same squad then Mr. Meeseeks number e should be in that same squad.Also, each squad needs its own presidio, and building a presidio needs money, so they want the total number of squads to be minimized.Rick and Morty haven't finalized the exact value of k, so in order to choose it, for each k between 1 and n (inclusive) need to know the minimum number of presidios needed.

Input

The first line of input contains a single integer n (1?≤?n?≤?105) — number of Mr. Meeseeks.The second line contains n integers a1,?a2,?...,?an separated by spaces (1?≤?ai?≤?n) — colors of Mr. Meeseeks in order they standing in a line.

Output

In the first and only line of input print n integers separated by spaces. i-th integer should be the minimum number of presidios needed if the value of k is i.

Examples
Input

5
1 3 4 3 3

Output

4 2 1 1 1 

Input

8
1 5 7 8 1 7 6 1

Output

8 4 3 2 1 1 1 1 

【題目分析】
這道題放在線段樹里面我實在沒有什么想法，覺得就暴力一下不可以嗎？可是估計了一下時間應該會超時，在網上看到一種想法，就是二分暴力，如果區間兩端答案一樣那么說明中間的所有的答案都是一樣的，因為答案應該是非線性遞減的。
【AC代碼】

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<climits>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;typedef long long ll;
const int INF=0x3f3f3f3f;
const int MAXN=1e5+5;
int a[MAXN];
int vis[MAXN];
int ans[MAXN];
int n;void Read()
{scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);}
}int GetAns(int k)	//模擬
{memset(vis,-1,sizeof(vis));int cnt=0,ret=1;for(int i=1;i<=n;i++){if(vis[a[i]]==ret) continue;cnt++;if(cnt>k){ret++;cnt=1;}vis[a[i]]=ret;}return ret;
}void Solve(int l,int r)	//二分
{if(l>r) return;int ansl=GetAns(l);int ansr=GetAns(r);if(ansl==ansr){for(int i=l;i<=r;i++){ans[i]=ansl;}return;}ans[l]=ansl; ans[r]=ansr;int mid=(l+r)>>1;Solve(l+1,mid); Solve(mid+1,r-1);
}void Print()
{for(int i=1;i<=n;i++){printf("%d ",ans[i]);}
}int main()
{Read();Solve(1,n);Print();return 0;
}