Stocking but not completely stocking (daily question 2 in spring)

A little-known fact is , Cows have their own words ：「 Niu Wen 」.

Niu Wenyou Letters ​​​a​​​ To ​​z​​​ form , But when the cow says Niu Wen , May be familiar to us ​​abcdefghijklmnopqrstuvwxyz​​ Different , She will arrange the letters in a certain order .

To pass the time , cow Bessie Humming the cow alphabet song repeatedly , and Farmer John Wonder how many times she sang .

Given a string of lowercase letters , by Farmer John hear Bessie Sing the letters , Calculation Bessie At least a few times I sang the complete song of the cow alphabet , bring Farmer John Can hear a given string .

Farmer John Not always pay attention to Bessie What you sing , So he may miss listening Bessie Some of the letters sung .

The given string contains only the letters he remembers he hears .

Input format
The first line of input contains Lowercase letters , by ​​​a​​​ To ​​z​​ The order of the Niu alphabet .

The next line contains a string of lowercase letters , by Farmer John hear Bessie Sing the letters .

Output format
Output Bessie The minimum number of complete niuwen alphabet songs sung .

Data range
The length of the string is not less than And not more than .

sample input ：

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       abcdefghijklmnopqrstuvwxyz
       
       
mood
      
      
      
      

       
       
1.
       
       
2.
      
      

sample output ：

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       3
      
      
      
      

       
       
1.
      
      

Sample explanation
In this example , The Niu alphabet is in line with the daily alphabet .

Bessie At least three times .

There may be Bessie I only sang the song of the cow alphabet three times , and Farmer John I heard the following capitalized letters .

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       abcdefghijklMnOpqrstuvwxyz
       
       
abcdefghijklmnOpqrstuvwxyz
       
       
abcDefghijklmnopqrstuvwxyz
      
      
      
      

       
       
1.
       
       
2.
       
       
3.
      
      

---

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       #include<iostream>
       
       

       
       #include<cstdio>
       
       

       
       #include<cstring>
       
       

       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       

       
       const 
       
       int 
       
       N 
       
       = 
       
       1010;
       
       

       
       

       
       char 
       
       a[
       
       30], 
       
       s[
       
       N];
       
       

       
       int 
       
       pos[
       
       30];
       
       

       
       

       
       int 
       
       main(){
       
       
    
       
       
    
       
       scanf(
       
       "%s", 
       
       a);
       
       
    
       
       for(
       
       int 
       
       i 
       
       = 
       
       0; 
       
       i 
       
       < 
       
       26; 
       
       i
       
       ++) 
       
       pos[
       
       a[
       
       i]
       
       -
       
       'a'] 
       
       = 
       
       i;
       
       
    
       
       
    
       
       scanf(
       
       "%s", 
       
       s);
       
       
    
       
       int 
       
       n 
       
       = 
       
       strlen(
       
       s);
       
       
    
       
       
    
       
       int 
       
       pre 
       
       = 
       
       30;
       
       
    
       
       int 
       
       res 
       
       = 
       
       0;
       
       
    
       
       for(
       
       int 
       
       i 
       
       = 
       
       0; 
       
       i 
       
       < 
       
       n; 
       
       i
       
       ++){
       
       
        
       
       
        
       
       int 
       
       x 
       
       = 
       
       pos[
       
       s[
       
       i]
       
       -
       
       'a'];
       
       
        
       
       if(
       
       x 
       
       <= 
       
       pre) 
       
       res
       
       ++;
       
       
        
       
       pre 
       
       = 
       
       x;
       
       
    }
       
       
    
       
       
    
       
       cout 
       
       << 
       
       res 
       
       << 
       
       endl;
       
       
    
       
       
    
       
       return 
       
       0;
       
       
}
      
      
      
      

       
       
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31.
       
       
32.