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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#include <math.h>


int main(void)
{
	long int i, j;
	long int a;
	long int sum;
	long int count;
	long int flg;
	long int getflg;
	long int alfs[26] = {0};
	long int alft[26] = {0};
	char s[100001];
	char t[100001];
	
	scanf("%s%s", s, t);
	
	for(i = 0; i < strlen(s); i++) {
		alfs[s[i] - 'a']++;
	}
	
	for(i = 0; i < strlen(t); i++) {
		alft[t[i] - 'a']++;
	}
	
	for(i = 0; i < 26; i++) {
		if( (alfs[i] == 0) && (alft[i] > 0) ) {
			printf("-1\n");
			return 0;
		}
	}
	
	//t:最終的にしたい文字 s:材料
	count = 0;
	flg = 0;
	for(i = 0; i < strlen(t); i++) {
		getflg = 0;
		for(j = 0; j < strlen(s); j++) {
			if(t[i] == s[j]) {
				i++;
				getflg = 1;
				if(i >= strlen(t)) {
					flg = 1;
					break;
				}
				continue;
			}
		}
		if(flg) {
			break;
		}
		if(getflg) {
			i--;
		}
		count++;
	}
	
	printf("%ld\n", count);
	
	sum = strlen(s) * count;
	
	printf("%ld\n", sum);
	
	for(i = 0; i < strlen(s); i++) {
		if(t[strlen(t) - 1] == s[i]) {
			sum += (i + 1);
			printf("%ld\n", i);
			break;
		}
	}
	
	printf("%ld\n", sum);
	
	return 0;
}