A string is finite sequence of characters over a non-empty finite set Σ.

In this problem, Σ is the set of lowercase letters.

Substring, also called factor, is a consecutive sequence of characters occurrences at least once in a string.

Now your task is a bit harder, for some given strings, find the length of the longest common substring of them.

Here common substring means a substring of two or more strings.

Input

The input contains at most 10 lines, each line consists of no more than 100000 lowercase letters, representing a string.

Output

The length of the longest common substring. If such string doesn't exist, print "0" instead.

Example

Input:
 

alsdfkjfjkdsal
fdjskalajfkdsla
aaaajfaaaa

Output:
2

之前用kmp寫過類似的題，估計這個鐵定超，，

在一個串上建sam，然後開始掃，lcs表示當前串匹配到這個結點時和sam上串的最大公共子串長度，nlcs表示前n個的

ac程式碼

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
#define N 300010 
struct sam
{
	sam *pre,*son[26];
	int len,lcs,nlcs;
}que[N],*root,*tail,*b[N];
int tot;
void add(int c,int l)
{
	sam *p=tail,*np=&que[tot++];
	np->len=np->nlcs=l;
	tail=np;
	while(p&&p->son[c]==NULL)
	{
		p->son[c]=np;
		p=p->pre;
	}
	if(p==NULL)
		np->pre=root;
	else
	{
		sam *q=p->son[c];
		if(p->len+1==q->len)
			np->pre=q;
		else
		{
			sam *nq=&que[tot++];
			*nq=*q;
			nq->len=nq->nlcs=p->len+1;
			np->pre=q->pre=nq;
			while(p&&p->son[c]==q)
			{
				p->son[c]=nq;
				p=p->pre;
			} 
		}
	}
}
char str[N>>1];
int cnt[N>>1],len;
void build()
{
	int i,j;
	root=tail=&que[tot++];
	for(i=0;i<len;i++)
		add(str[i]-'a',i+1);
	memset(cnt,0,sizeof(cnt));
	for(i=0;i<tot;i++)
		cnt[que[i].len]++;
	for(i=1;i<=len;i++)
		cnt[i]+=cnt[i-1];
	for(i=0;i<tot;i++)
		b[--cnt[que[i].len]]=&que[i];
}
int main()
{
	int i;
	scanf("%s",str);
	len=strlen(str);
	build();
	while(scanf("%s",str)!=EOF)
	{
		sam *p=root;
		int l=0,i;
		for(i=0;str[i];i++)
		{
			int now=str[i]-'a';
			if(p->son[now])
			{
				l++;
				p=p->son[now];
			}
			else
			{
				while(p&&p->son[now]==NULL)
				{
					p=p->pre;
				}
				if(p==NULL)
				{
					l=0;
					p=root;
				}
				else
				{
					l=p->len+1;
					p=p->son[now];
				}
			}
			if(l>p->lcs)
				p->lcs=l;
		}
		for(i=tot-1;i>=0;i--)
		{
			p=b[i];
			if(p->lcs<p->nlcs)
				p->nlcs=p->lcs;
			if(p->pre&&p->pre->lcs<p->lcs)
				p->pre->lcs=p->lcs;
			p->lcs=0;
		}
	}
	int ans=0;
	for(i=0;i<tot;i++)
	{
		if(que[i].nlcs>ans)
			ans=que[i].nlcs;
	}
	printf("%d\n",ans);
}