Codeforces 91A-Newspaper Headline

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本文探讨了一种算法问题，即如何通过拼接特定数量的原始字符串并从中删除部分字符来形成目标字符串。该问题涉及字符串操作、字符匹配等关键技术，并提供了一个具体的实现示例。

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A newspaper is published in Walrusland. Its heading is s1, it consists of lowercase Latin letters. Fangy the little walrus wants to buy several such newspapers, cut out their headings, glue them one to another in order to get one big string. After that walrus erase several letters from this string in order to get a new word s2. It is considered that when Fangy erases some letter, there's no whitespace formed instead of the letter. That is, the string remains unbroken and it still only consists of lowercase Latin letters.

For example, the heading is "abc". If we take two such headings and glue them one to the other one, we get "abcabc". If we erase the letters on positions 1 and 5, we get a word "bcac".

Which least number of newspaper headings s1 will Fangy need to glue them, erase several letters and get word s2?

Input

The input data contain two lines. The first line contain the heading s1, the second line contains the word s2. The lines only consist of lowercase Latin letters (1 ≤ |s1| ≤ 104, 1 ≤ |s2| ≤ 106).

Output

If it is impossible to get the word s2 in the above-described manner, print "-1" (without the quotes). Otherwise, print the least number of newspaper headings s1, which Fangy will need to receive the word s2.

Examples

input
abc
xyz

output
-1

input
abcd
dabc

output
2

题意：给定一个A串，一个B串，用不同的去除某些位置的A串组成B串，问最少需要几串？

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>

using namespace std;

char s1[10009],s2[1000009];
int a[10009][26],b[26];

int main()
{
    while(~scanf("%s %s",s1,s2))
    {
        memset(a,-1,sizeof(a));
        memset(b,0,sizeof b);
        int sum=0,flag=0;
        int len1=strlen(s1),len2=strlen(s2);
        for(int i=0; i<len1; i++)
        {
            b[s1[i]-'a']=1;
            for (int j=0; j<=i; j++)
                if (a[j][s1[i]-'a']==-1) a[j][s1[i]-'a']=i;
        }
        for (int i=0; i<len2; i++)
            if (!b[s2[i]-'a'])
            {
                printf("-1\n");
                flag=1;break;
            }
        if(flag) continue;
        int j=0;
        while(j<len2)
        {
            sum++;
            int i=0;
            while(j<len2&&i<len1&&a[i][s2[j]-'a']!=-1)
            {
                i=a[i][s2[j]-'a']+1;
                j++;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}