## Longest Common Substring

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 simple, for two 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 exactly two lines, each line consists of no more than 250000 lowercase letters, representing a string.

### Output

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

### Example

`Input:alsdfkjfjkdsalfdjskalajfkdslaOutput:3求兩個字符串的最大公共子串的長度`

`#include <iostream>#include <string.h>#include <algorithm>#include <stdio.h>using namespace std;const int N=250005;struct State{    State *link,*go[26];    int step;    void clear()    {        link=0;        step=0;        memset(go,0,sizeof(go));    }}*root,*last;State statePool[N*2],*cur;void init(){    cur=statePool;    root=last=cur++;    root->clear();}void Insert(int w){    State *p=last;    State *np=cur++;    np->clear();    np->step=p->step+1;    while(p&&!p->go[w])        p->go[w]=np,p=p->link;    if(p==0)        np->link=root;    else    {        State *q=p->go[w];        if(p->step+1==q->step)            np->link=q;        else        {            State *nq=cur++;            nq->clear();            memcpy(nq->go,q->go,sizeof(q->go));            nq->step=p->step+1;            nq->link=q->link;            q->link=nq;            np->link=nq;            while(p&&p->go[w]==q)                p->go[w]=nq, p=p->link;        }    }    last=np;}char A[N],B[N];int main(){    int maxx,len;    while(scanf("%s%s",A,B)!=EOF)    {        len=0;        maxx=0;        int n,m;        n=strlen(A);        m=strlen(B);        init();        State *p=root;        for(int i=0;i<n;i++)         Insert(A[i]-'a');       for(int i=0;i<m;i++)       {           if(p->go[B[i]-'a'])           {               len++;               p=p->go[B[i]-'a'];           }           else           {             while(p&&!p->go[B[i]-'a'])p=p->link;             if(!p)             {              p=root;              len=0;             }             else             {                 len=p->step+1;                 p=p->go[B[i]-'a'];             }           }           maxx=max(maxx,len);        }        cout<<maxx<<endl;    }    return 0;}`