D. Match & Catch 后缀数组

Posted 2020-09-04 Heilce

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.

Examples

Input

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?=?a₁a₂a₃...a_|a|, where |a| is the length of string a, and a_i is the i^th letter of the string.

We will call string a_la_l?+?1a_l?+?2...a_r (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 l₁,?r₁ such that l₁?≠?l and the substring [l₁,?r₁] is equal to the substring [l,?r] in a.

 

好长时间没有写过后缀数组的问题了，看了一下午的课本回顾了一下。。。。

这个题就是属于后缀数组的基础的问题，感觉就是套模板。下面贴一下后缀数组的模板。

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;
        //基数排序第一关键字
        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;

    }
}