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: Wrong Answer)

#include <stdio.h>
#define MXSZ 100010
char s[MXSZ],t[MXSZ];
int P[27][MXSZ],C[27];
void init(){
}
int bSearch(int *ar,int l,int r,int v){
    int ret = -1;
    while(l<=r){
        int m = (l+r)/2;
        if(ar[m]>=v){
            ret = ar[m];
            r = m-1;
        }
        else l = m+1;
    }
    return ret;
}
void MRoy(){
    init();
    scanf("%s%s",s,t);
    long long L = 0;
    for(int i = 0;s[i]!='\0';i++){
        int T = s[i]-'a';
        P[T][C[T]++] = i;
        L++;
    }
    long long int Loop = 0;int pos = -1;
    for(int i = 0;t[i]!='\0';i++){
        int T = t[i]-'a';
        if(C[T]==0) {
            printf("-1\n");
            return;
        }
        int res = bSearch(P[T],0,C[T]-1,pos+1);
        if(res==-1) Loop++;
        pos = P[T][0];
    }
    printf("%lld\n",Loop*L+pos+1ll);
    return;
}
int main(){
    MRoy();
    return 0;
}