D. Match & Catch 尾碼數組

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Police headquarter is monitoring signal on different frequency levels. They have got two suspiciously encoded strings s_{1 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 s_{1 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 s_{1 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 s_{1 and s2. If there are no common unique substrings of s1 and s2 print -1.}

ExamplesInput

apple
pepperoni

Output

2

Input

lover
driver

Output

1

Input

bidhan
roy

Output

-1

Input

testsetses
teeptes

Output

3

Note

Imagine we have string a?=?a_{1a2a3...a|a|, where |a| is the length of string a, and ai is the i^{th letter of the string.}}

We will call string a_{lal?+?1al?+?2...ar (1?≤?l?≤?r?≤?|a|) the substring [l,?r] of the string a.}

The substring [l,?r] is unique in a if and only if there is no pair l_{1,?r1 such that l1?≠?l and the substring [l1,?r1] is equal to the substring [l,?r] in a.}

 

好長時間沒有寫過尾碼數組的問題了，看了一下午的課本回顧了一下。。。。

這個題就是屬於尾碼數組的基礎的問題，感覺就是套模板。下面貼一下尾碼數組的模板。

const int maxn=1e4+5;int t[maxn],t2[maxn];int sa[maxn],c[maxn];char s[maxn];int n;void build(int m){    int *x=t,*y=t2;    for(int i=0;i<m;i++) c[i]=0;    for(int i=0;i<n;i++) c[x[i]=s[i]]++;    for(int i=1;i<m;i++) c[i]+=c[i-1];    for(int i=n-1;i>=0;i--) sa[--c[x[i]]]=i;    for(int k=1;k<=n;k<<=1)    {        int p=0;        for(int i=n-k;i<n;i++) y[p++]=i;        for(int i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;        //基數排序第一關鍵字        for(int i=0;i<m;i++) c[i]=0;        for(int i=0;i<n;i++) c[x[y[i]]]++;        for(int i=1;i<m;i++) c[i]+=c[i-1];        for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];        swap(x,y);        p=1; x[sa[0]]=0;        for(int i=1;i<n;i++) x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;        if(p>=n) break;        m=p;    }}int rank[maxn],height[maxn];void getheight(){    int k=0;    for(int i=0;i<n;i++) rank[sa[i]]=i;    for(int i=0;i<n;i++)    {        if(k) k--;        if(rank[i]==0)        {            height[rank[i]]=0;            continue;        }        int j=sa[rank[i]-1];        while(s[i+k]==s[j+k]) k++;        height[rank[i]]=k;    }}

 

D. Match & Catch 尾碼數組