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>
#include <stdlib.h>
#include <string.h>

int main(void){
    int i, j, a[27], b[27], p, q, M, m;
    int *A;
    A = malloc(sizeof(int) * 5500000);
    long long X;
    char s[200005], t[200005];
    
    X=0; p=0; q=0;
    scanf("%s", s);
    scanf("%s", t);
    
    for (i=1; i<=26; i++) {
        a[i]=0; b[i]=0;
    }
    for (i=0; i<strlen(s); i++) {
        a[s[i]-96]++;
        A[200000*(s[i]-96)+(a[s[i]-96])]=i;
        // printf("(%d, %d)\n", 200000*(s[i]-96)+(a[s[i]-96]), i);
    }
    for (i=0; i<strlen(t); i++) {
        /*
        printf("[%d]\n", i);
        for (j=1; j<=a[t[i]-96]; j++) printf("%d, ", A[200000*(t[i]-96)+(j)]);
        printf("\n");
        */
        b[t[i]-96]++;
    }
    for (i=1; i<=26; i++) {
        if (a[i]==0 && b[i]!=0) {
            printf("%d", -1);
            return 0;
        }
    };
    
    for (i=0; i<strlen(t); i++) {
        m=1000000;
        for (j=1; j<=a[t[i]-96]; j++) {
            M=0;
            if (q>=A[200000*(t[i]-96)+(j)]) M+=strlen(s);
            M+=A[200000*(t[i]-96)+(j)];
            if (M<m) m=M;
        }
        if (m>=strlen(s)) {
            p++;
        }
        q=m%strlen(s);
        // printf("(%d, %d, %d)\n", i, p, q);
    }
    printf("%lld", (long long)p*(long long)strlen(s)+(long long)q+(long long)1);
    
    return 0;
}