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

Notice: new testcases added

第一个串建SAM,剩下的串在上面跑....记录下每个节点的最小匹配长度(初始值是 len 即能表示的最长后缀),

从后向前更新fa节点的LCS....

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=110000;

struct SAM_Node
{
    SAM_Node *fa,*next[26];
    int len,id,pos;
    SAM_Node(){}
    SAM_Node(int _len)
    {
        fa=0; len=_len;
        memset(next,0,sizeof(next));
    }
};

SAM_Node SAM_node[maxn*2],*SAM_root,*SAM_last;
int SAM_size;

SAM_Node *newSAM_Node(int len)
{
    SAM_node[SAM_size]=SAM_Node(len);
    SAM_node[SAM_size].id=SAM_size;
    return &SAM_node[SAM_size++];
}

SAM_Node *newSAM_Node(SAM_Node *p)
{
    SAM_node[SAM_size]=*p;
    SAM_node[SAM_size].id=SAM_size;
    return &SAM_node[SAM_size++];
}

void SAM_init()
{
    SAM_size=0;
    SAM_root=SAM_last=newSAM_Node(0);
    SAM_node[0].pos=0;
}

void SAM_add(int x,int len)
{
    SAM_Node *p=SAM_last,*np=newSAM_Node(p->len+1);
    np->pos=len; SAM_last=np;
    for(;p&&!p->next[x];p=p->fa)
        p->next[x]=np;
    if(!p)
    {
        np->fa=SAM_root;
        return ;
    }
    SAM_Node *q=p->next[x];
    if(q->len==p->len+1)
    {
        np->fa=q;
        return ;
    }
    SAM_Node *nq=newSAM_Node(q);
    nq->len=p->len+1;
    q->fa=nq; np->fa=nq;
    for(;p&&p->next[x]==q;p=p->fa)
        p->next[x]=nq;
}

char str[maxn],other[maxn];
int LCS[maxn*2],len,n;
int c[maxn*2]; SAM_Node *top[maxn*2];

int main()
{
    scanf("%s",str);
    len=strlen(str);
    SAM_init();
    for(int i=0;i<len;i++)
        SAM_add(str[i]-'a',i+1);

    for(int i=0;i<SAM_size;i++)
        c[SAM_node[i].len]++;
    for(int i=1;i<=len;i++)
        c[i]+=c[i-1];
    for(int i=0;i<SAM_size;i++)
        top[--c[SAM_node[i].len]]=&SAM_node[i];

    while(scanf("%s",other)!=EOF)
    {
        n++;
        int len2=strlen(other);
        SAM_Node* now=SAM_root;
        int temp=0;
        for(int i=0;i<len2;i++)
        {
            int x=other[i]-'a';
            if(now->next[x])
            {
                temp++;
                now=now->next[x];
            }
            else
            {
                while(now&&now->next[x]==0)
                    now=now->fa;
                if(now)
                {
                    temp=now->len+1;
                    now=now->next[x];
                }
                else
                {
                    temp=0;
                    now=SAM_root;
                }
            }
            if(temp>LCS[now->id])
                LCS[now->id]=temp;
        }
        for(int i=SAM_size-1;i>=0;i--)
        {
            if(LCS[top[i]->id]<top[i]->len)
                top[i]->len=LCS[top[i]->id];
            if(top[i]->fa&&LCS[top[i]->fa->id]<LCS[top[i]->id])
               LCS[top[i]->fa->id]=LCS[top[i]->id];
            LCS[top[i]->id]=0;
        }
    }
    int ans=0;
    for(int i=1;i<SAM_size;i++)
    {
        if(ans<SAM_node[i].len)
            ans=SAM_node[i].len;
    }
    printf("%d\n",ans);
    return 0;
}

SPOJ LCS2 1812. Longest Common Substring II