The next "Data Structures and Algorithms" lesson will be about Longest Increasing Subsequence (LIS for short) of a sequence. For better understanding, Nam decided to learn it a few days before the lesson.

Nam created a sequence a consisting of n (1 ≤ n ≤ 105) elements a1, a2, ..., an (1 ≤ ai ≤ 105). A subsequence ai1, ai2, ..., aik where 1 ≤ i1 < i2 < ... < ik ≤ n is called increasing if ai1 < ai2 < ai3 < ... < aik. An increasing subsequence is called longest if it has maximum length among all increasing subsequences.

Nam realizes that a sequence may have several longest increasing subsequences. Hence, he divides all indexes i (1 ≤ i ≤ n), into three groups:

group of all i such that ai belongs to no longest increasing subsequences.
    group of all i such that ai belongs to at least one but not every longest increasing subsequence.
    group of all i such that ai belongs to every longest increasing subsequence.

Since the number of longest increasing subsequences of a may be very large, categorizing process is very difficult. Your task is to help him finish this job.

#include<iostream>
#include<stdio.h>
#include<map>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 100005
int b[maxn];
int a[maxn];
void add(int x,int val)
{while(x<=100000){b[x] = max(b[x],val);x += x & (-x);}
}
int get(int x)
{int ans = 0;while(x){ans = max(ans,b[x]);x -= x & (-x);}return ans;
}
int dp1[maxn];
int dp2[maxn];
int ans[maxn];
map<int,int> H;
int main()
{int n;scanf("%d",&n);int LIS = 0;for(int i=1;i<=n;i++){scanf("%d",&a[i]);dp1[i] = 1 + get(a[i]-1);add(a[i],dp1[i]);LIS = max(LIS,dp1[i]);}reverse(a+1,a+1+n);memset(b,0,sizeof(b));for(int i=1;i<=n;i++){a[i] = 100000 - a[i] + 1;dp2[i] = 1 + get(a[i] - 1);add(a[i],dp2[i]);}reverse(dp2+1,dp2+1+n);for(int i=1;i<=n;i++){if(dp1[i]+dp2[i]-1!=LIS)ans[i]=1;else H[dp1[i]]++;}for(int i=1;i<=n;i++){if(ans[i]!=1&&H[dp1[i]]==1){ans[i]=3;}}for(int i=1;i<=n;i++)if(ans[i]==1)cout<<"1";else if(ans[i]==0)cout<<"2";else if(ans[i]==3)cout<<"3";
}
/*
10
2 2 2 17 8 9 10 17 10 5
*/