Problem:
Score : 500 points

Problem Statement

Given are two strings s and t consisting of lowercase English letters. Determine if there exists an integer i satisfying the following condition, and find the minimum such i if it exists.

Let s' be the concatenation of 10^{100} copies of s. t is a subsequence of the string {s'}_1{s'}_2\ldots{s'}_i (the first i characters in s').

Notes

A subsequence of a string a is a string obtained by deleting zero or more characters from a and concatenating the remaining characters without changing the relative order. For example, the subsequences of contest include net, c, and contest.

Constraints

1 \leq |s| \leq 10^5

1 \leq |t| \leq 10^5

s and t consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s
t

Output

If there exists an integer i satisfying the following condition, print the minimum such i; otherwise, print -1.

Sample Input 1

contest
son

Sample Output 1

10

t = son is a subsequence of the string contestcon (the first 10 characters in s' = contestcontestcontest...), so i = 10 satisfies the condition.

On the other hand, t is not a subsequence of the string contestco (the first 9 characters in s'), so i = 9 does not satisfy the condition.

Similarly, any integer less than 9 does not satisfy the condition, either. Thus, the minimum integer i satisfying the condition is 10.

Sample Input 2

contest
programming

Sample Output 2

-1

t = programming is not a substring of s' = contestcontestcontest.... Thus, there is no integer i satisfying the condition.

Sample Input 3

contest
sentence

Sample Output 3

33

Note that the answer may not fit into a 32-bit integer type, though we cannot put such a case here.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int d[100010][30],n;
int main(){
  char s[100010],t[100010];
  long long ans=0,a,l,i,b[30];
  scanf("%s %s",s,t);
  for(l=0;s[l];l++);
  for(i=0;i<30;i++)b[i]=-1;
  for(i=l;i--;)b[s[i]-'a']=i+l;
  for(a=l;a--;){
    for(i=0;i<30;i++)d[a][i]=b[i];
    b[s[a]-'a']=a;
  }
  ans=a=b[t[0]-'a'];//printf("%lld\n",ans);
  if(ans<0)goto END;
  for(i=1;t[i];i++){//printf("%d\n",d[a][t[i+1]+'a']);
    n=d[a][t[i]-'a'];//printf("%d\n",n);
    if(n<0)goto END;
    ans+=n-a;
    a=n%l;
  }
  printf("%lld\n",ans+1);
  return 0;
 END:;
  printf("-1\n");
  return 0;
}