D. Match & Catch suffix array

Police headquarter is monitoring signal on different frequency levels. They has got, suspiciously encoded strings s_{1 and s2 from the different frequencies As signals. They is suspecting that these, strings is from, different criminals and they is planning to do some evil task.}

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

Given strings s_{1 and s2 consist of lowercase Latin letters, find the smallest (by Le ngth) Common substring p of both s1 and s2, where p I s 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?≤?| s 1|,?| s 2|? ≤?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 is 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 am the i^{th Letter of the string.}}

We'll call string a _{l a l ? +?1 a l ? +?2 ... A R (1?≤? L ? ≤? r ? ≤?| a |) The substring [ l , R ] of the string a .}

The substring [l,? R] is the 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.}

For a long time did not write after the problem of the array, read the afternoon of the textbook review a bit ....

This problem is the base of the suffix array, the feeling is a set of templates. Paste the suffix array template below.

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;        Cardinal sort first keyword 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 suffix array