CodeForces-427D：Match & Catch （后缀自动机）

Police headquarter is monitoring signal on different frequency levels. They have got two suspiciously encoded strings s₁ and s₂ from two different frequencies as signals. They are suspecting that these two strings are from two different criminals and they are planning to do some evil task.

Now they are trying to find a common substring of minimum length between these two strings. The substring must occur only once in the first string, and also it must occur only once in the second string.

Given two strings s₁ and s₂ consist of lowercase Latin letters, find the smallest (by length) common substring p of both s₁ and s₂, where p is a unique substring in s₁ and also in s₂. See notes for formal definition of substring and uniqueness.

Input

The first line of input contains s₁ and the second line contains s₂ (1 ≤ |s₁|, |s₂| ≤ 5000). Both strings consist of lowercase Latin letters.

Output

Print the length of the smallest common unique substring of s₁ and s₂. If there are no common unique substrings of s₁ and s₂ print -1.

Example

Input

apple
pepperoni

Output

Input

lover
driver

Output

Input

bidhan
roy

Output

-1

Input

testsetses
teeptes

Output

题意：给定字符串S，T。求最短的公共字串s的长度|s|，s满足在各自的母串里只出现了一次。

思路：建立S的后缀自动机，拓扑得到每个字串出现的次数num1。然后T在后缀自动机上面走，也得到字串T的次数num2。然后每个集合如果在各自串都只出现一次，就可以用当前集合的最小长度比较答案了。

经验：对建立后缀自动机的S串，在设置num1的时候，可以在np处设置num1=1（也就是所有Blue节点），或者建立完自动机后跑一遍，跑到的地方num1=1。

具体的可以参考代码。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int inf=10000000;
const int maxn=20005;
int ans=inf,L; char c[maxn];    
struct Sam
{
    int fa[maxn],ch[maxn][26],maxlen[maxn],Last,cnt;
    int a[maxn],b[maxn],num1[maxn],num2[maxn],np,nq,p,q;
    Sam(){
        Last=cnt=1;
    }
    void add(int x)
    {
        np=++cnt; p=Last; Last=np;
        maxlen[np]=maxlen[p]+1; num1[np]++;//方法1 
        while(p&&!ch[p][x]) ch[p][x]=np,p=fa[p];
        if(!p) fa[np]=1;
        else {
            q=ch[p][x];
            if(maxlen[p]+1==maxlen[q]) fa[np]=q;
            else {
                nq=++cnt; maxlen[nq]=maxlen[p]+1;
                fa[nq]=fa[q]; fa[q]=fa[np]=nq;
                memcpy(ch[nq],ch[q],sizeof(ch[nq]));
                while(p&&ch[p][x]==q) ch[p][x]=nq,p=fa[p];
            }
        }
    }
    void sort()
    {
        for(int i=1;i<=cnt;i++) a[maxlen[i]]++;
        for(int i=1;i<=cnt;i++) a[i]+=a[i-1];
        for(int i=1;i<=cnt;i++) b[a[maxlen[i]]--]=i;    
        //for(int i=1,tp=1;i<=cnt;i++){
        //    tp=ch[tp][c[i]-'a']; num1[tp]++;
        //}
        //方法2 
        for(int i=cnt;i>=0;i--) num1[fa[b[i]]]+=num1[b[i]];
    }
    void solve()
    {
        scanf("%s",c+1); L=strlen(c+1); p=1;
        for(int i=1;i<=L;i++){
            int x=c[i]-'a';
            if(ch[p][x]) p=ch[p][x];
            else {
                while(p&&!ch[p][x]) p=fa[p];
                if(!p) p=1;
                else p=ch[p][x];
            }
            num2[p]++;
        }
        for(int i=cnt;i>=0;i--) num2[fa[b[i]]]+=num2[b[i]];
        for(int i=2;i<=cnt;i++) 
          if(num1[i]==1&&num2[i]==1)
           ans=min(ans,maxlen[fa[i]]+1);
        if(ans==inf) ans=-1; 
    }
}sam;
int main()
{
    scanf("%s",c+1); L=strlen(c+1);
    for(int i=1;i<=L;i++) sam.add(c[i]-'a');
    sam.sort(); sam.solve();
    printf("%d
",ans);
    return 0;
}