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 Python (3.4.3) (judge verdict: Accepted)

#!/usr/bin/env python3
import sys
import bisect

def solve(s: str, t: str):
    LENS = len(s)
    dict_s = {} ##{'c': [0], 'o': [1], 'n': [2], 't': [3, 6], 'e': [4], 's': [5]}
    index = [] ##[[5], [4], [2], [3, 6], [4], [2], [0], [4]]

    ## create dict
    for i in range(LENS):
        if dict_s.get(s[i]) != None:
            dict_s[s[i]].append(i)
        else:
            dict_s[s[i]]= [i]
    
    ## そもそもtが作れるかの判定
    for tt in t:
        if dict_s.get(tt) == None:
            print(-1)
            return
        else:
            index.append(dict_s[tt])
    
    answer = 0
    for i in range(len(index)):
        cur = index[i]

        if i == 0:
            answer += cur[0]
            index[i]=[cur[0]]
            continue

        prev = index[i-1]

        if len(cur)==1:
            if cur[0]>prev[0]:
                answer += cur[0]-prev[0]
            else:
                answer += cur[0]+LENS-prev[0] ##一周追加
        else: ##複数
            next_index = bisect.bisect_left(cur,prev[0])

            if next_index<len(cur)-1: ## prevより大きいindexがある時
                next_index += 1 if cur[next_index] == prev[0] else 0
                
                answer += cur[next_index]-prev[0]
                index[i]= [cur[next_index]]
            elif next_index==len(cur)-1:
                if prev[0]<cur[next_index]:
                    answer += cur[next_index]-prev[0]
                    index[i] =[cur[next_index]]
                elif prev[0]==cur[next_index]:
                    answer += cur[0]+LENS-prev[0]
                    index[i] =[cur[0]]
            else: ##next_index==len(cur):
                answer += cur[0]+LENS-prev[0]
                index[i] = [cur[0]]
       

    print(answer+1)
    return


def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    s = next(tokens)  # type: str
    t = next(tokens)  # type: str
    solve(s, t)

if __name__ == '__main__':
    main()