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:
alsdfkjfjkdsal
fdjskalajfkdsla

Output:
3

n四次方的暴力不解釋 = =
然後還有n方的dp
然後就是sa和sam
sa的復雜度是nlogn
然後就是sam了，玄學的復雜，完全不知道如何證明，o（玄學)就對了

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #include<map>
 5 #include<set>
 6 #include<queue>
 7
 
 #include<algorithm>
 8 #include<cmath>
 9 #include<stack>
10 #include<cstring>
11 #define mem(a,b) memset(a,b,sizeof a)
12 #define sc(a) scanf("%d",&(a))
13 #define scc(a,b) scanf("%d %d",&(a),&(b))
14 #define ll long long
15 using namespace std;
16 
17 
18 int kd[1000000];
19 
20 struct dd
21 {
22 
23     int fa,len;
24     int tr[29];
25 
26 }dian[1200000];
27 int cnt[1200000];
28 
29 int last=1;int tot=1;
30 inline void add(int c)
31 {
32    int p=last; int np=last=++tot;
33    dian[np].len=dian[p].len+1;
34 
35    for(;p&&!dian[p].tr[c];p=dian[p].fa)dian[p].tr[c]=np;
36    if(!p)dian[np].fa=1,cnt[1]++;
37    else
38    {
39        int q=dian[p].tr[c];
40        if(dian[q].len==dian[p].len+1)dian[np].fa=q,cnt[q]++;
41        else
42        {
43            int nq=++tot;
44            dian[nq]=dian[q];
45            dian[nq].len=dian[p].len+1;
46            dian[q].fa=dian[np].fa=nq;
47            cnt[nq]+=2;
48            for(;p&&dian[p].tr[c]==q;p=dian[p].fa)dian[p].tr[c]=nq;
49 
50        }
51    }
52 }
53 int ans=0;
54 void top()
55 {
56    int now=1,t=0;
57    string s;cin>>s; int len = s.length();
58    for(int i=0;i<len;i++)
59    {
60        int so=s[i]-‘a‘;
61        if(dian[now].tr[so])now=dian[now].tr[so],t++;
62        else
63        {
64            while(now&&!dian[now].tr[so])now=dian[now].fa;
65            if(!now)now=1,t=0;
66            else{
67 
68             t=dian[now].len+1;
69             //這裏先對t賦值，因為這裏是順著fa向上走，到達的狀態一定是兩個串
70             //的公共字串，然後多了s【i】，t+=1;
71             now=dian[now].tr[so];
72            }
73        }
74        ans=max(ans,t);
75    }
76    cout<<ans;
77 }
78 
79 signed main()
80 {
81     string  s; string t;
82     //ios::sync_with_stdio(0);
83 //    cin.tie(0);
84 
85     cin>>t;
86      s+=t;  int n=s.length();
87      for(int i=0;i<n;i++)add(s[i]-‘a‘);
88     top();
89 
90 
91 }

Longest Common Substring（後綴自動機）