You are going to take the entrance examination of Kyoto University tomorrow and have decided to memorize a set of strings S that is expected to appear in the examination. Since it is really tough to memorize S as it is, you have decided to memorize a single string T that efficiently contains all the strings in S.

You have already confirmed the following conditions for S and T are satisfied.

* Every string in S is a consecutive subsequence(1) in T .
* For every pair of strings x, y (x \neq y) in S , x is not a subsequence(2) of y.



Note that (1) is ''consecutive subsequence'', while (2) is ''subsequence''.

The next day, you opened the problem booklet at the examination and found that you can get a full score only if you remember S. However, You have forgot how to reconstruct S from T . Since all you remember is that T satisfies the above conditions, first you have decided to find the maximum possible number of elements in S .

Constraints

* 1 \leq |T| \leq 10^5
* T consists of only lowercase letters.



Partial points

* 30 points will be awarded for passing the test set satisfying the condition: 1 \leq |T| \leq 50 .
* Another 30 points will be awarded for passing the test set satisfying the condition: 1 \leq |T| \leq 10^3.

Input

The input is given from Standard Input in the following format:


T

The input only consists of T on one line.

Output

Print the maximum number of elements inS .

Examples

Input

abcabc


Output

3


Input

abracadabra


Output

7


Input

abcbabbcabbc


Output

8


Input

bbcacbcbcabbabacccbbcacbaaababbacabaaccbccabcaabba


Output

44
#include<iostream>
using namespace std;
//construct SA by SA-IS O(N)
#include<string>
#include<vector>
struct SA{
	string s;
	vector<int>sa;
	SA(const string&s_):s(s_)
	{
		sa=build(vector<int>(s.begin(),s.end()),256);
	}
	vector<int>induced_sort(const vector<int>&S,const vector<int>&id,const vector<bool>&SL,vector<int>last)
	{
		vector<int>first(last);
		vector<int>ret(id.size());
		ret[0]=id[0];
		for(int i=0;i<id.size();i++)
		{
			if(id[i]>=1&&!SL[id[i]-1])
			{
				ret[first[S[id[i]-1]-1]++]=id[i]-1;
			}
			else if(ret[i]>=1&&!SL[ret[i]-1])
			{
				ret[first[S[ret[i]-1]-1]++]=ret[i]-1;
			}
		}
		for(int i=id.size();i--;)
		{
			if(ret[i]>=1&&SL[ret[i]-1])
			{
				ret[--last[S[ret[i]-1]]]=ret[i]-1;
			}
		}
		return ret;
	}
	vector<int>build(vector<int>S,int maxval)
	{
		if(S.size()<=1)
		{
			return S.empty()?(vector<int>){0}:(vector<int>){1,0};
		}
		S.push_back(0);
		vector<int>cnt(maxval,0);
		vector<bool>SL(S.size());//S=>true,L=>false
		for(int i=S.size();i--;)
		{
			cnt[S[i]]+=1;
			SL[i]=i+1==S.size()||S[i]<S[i+1]||S[i]==S[i+1]&&SL[i+1];
		}
		for(int i=1;i<maxval;i++)cnt[i]+=cnt[i-1];
		vector<int>last(cnt);
		vector<int>id(S.size());
		vector<int>is_LMS(S.size());
		int LMScnt=0;
		for(int i=1;i<S.size();i++)
		{
			if(!SL[i-1]&&SL[i])
			{
				is_LMS[i]=1;
				id[--cnt[S[i]]]=i;
				LMScnt+=1;
			}
		}
		id=induced_sort(S,id,SL,last);
		int LMSsub=1;
		int pre=-1;
		is_LMS[id[0]]=LMSsub++;
		for(int i=1;i<id.size();i++)
		{
			if(is_LMS[id[i]])
			{
				if(pre>=0&&S[pre]==S[id[i]])
				{
					int k;
					for(k=1;S[pre+k]==S[id[i]+k]&&!is_LMS[pre+k]&&!is_LMS[id[i]+k];k++);
					LMSsub-=S[pre+k]==S[id[i]+k]&&is_LMS[pre+k]&&is_LMS[id[i]+k];
				}
				pre=id[i];
				is_LMS[id[i]]=LMSsub++;
			}
		}
		vector<int>newstr(LMScnt);
		vector<int>rev(LMScnt);
		int counter=0;
		for(int i=0;i<S.size();i++)
		{
			if(is_LMS[i])
			{
				newstr[counter]=is_LMS[i];
				rev[counter]=i;
				counter+=1;
			}
		}
		vector<int>sortedLMS=build(newstr,LMSsub);
		id.assign(S.size(),0);
		for(int i=1;i<sortedLMS.size();i++)
		{
			int I=rev[sortedLMS[i]];
			id[cnt[S[I]]++]=I;
		}
		return induced_sort(S,id,SL,last);
	}
	int operator[](int i)const{return sa[i];}
	int lower_bound(const string&t)const
	{
		int L=-1,R=sa.size();
		while(R-L>1)
		{
			int M=L+R>>1;
			if(s.compare(sa[M],t.size(),t)>=0)R=M;
			else L=M;
		}
		return R;
	}
	int upper_bound(const string&t)const
	{
		int L=-1,R=sa.size();
		while(R-L>1)
		{
			int M=L+R>>1;
			if(s.compare(sa[M],t.size(),t)<=0)L=M;
			else R=M;
		}
		return R;
	}
	bool contain(const string&t)const
	{
		int id=lower_bound(t);
		return id<sa.size()&&s.compare(sa[id],t.size(),t)==0;
	}
	int size()const{return sa.size();}
};
//construct LCP O(N) require SA
struct LCP{
	vector<int>lcp;
	LCP(const SA&sa)
	{
		int n=sa.size()-1;
		lcp.assign(n,0);
		vector<int>rank(n+1);
		for(int i=0;i<=n;i++)rank[sa[i]]=i;
		int h=0;
		lcp[0]=0;
		for(int i=0;i<n;i++)
		{
			int j=sa[rank[i]-1];
			h-=h>0;
			for(;i+h<n&&j+h<n&&sa.s[i+h]==sa.s[j+h];h++);
			lcp[rank[i]-1]=h;
		}
	}
	int operator[](int i)const{return lcp[i];}
	int size()const{return lcp.size();}
};
string s;
int f(int k,const SA&P,const LCP&Q)
{
	int cnt=0;
	int pre=0;
	for(int i=0;i<P.size();i++)
	{
		if(P[i]+k<=s.size()&&pre<k)
		{
			cnt++;
			pre=k;
		}
		if(i<Q.size()&&pre>Q[i])pre=Q[i];
	}
	return cnt;
}
int main()
{
	cin>>s;
	SA P(s);
	LCP Q(P);
	int L=1,R=s.size()+1;
	while(R-L>2)
	{
		int M1=(L+L+R)/3,M2=(L+R+R)/3;
		if(f(M1,P,Q)>f(M2,P,Q))R=M2;
		else L=M1;
	}
	int X=f(L,P,Q),Y=f(L+1,P,Q),Z=f(L+2,P,Q);
	if(X<Y)X=Y;
	if(X<Z)X=Z;
	cout<<X<<endl;
}
