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)

import sys
import bisect

sys.setrecursionlimit(10 ** 7)
f_inf = float('inf')
mod = 10 ** 9 + 7
alphabet = [chr(i) for i in range(97, 97 + 26)]


def resolve():
    S = input()
    T = input()

    n = len(S)
    m = len(T)

    idx = [[] for _ in range(26)]
    for i in range(n):
        idx[alphabet.index(S[i])].append(i)

    for i in range(n):
        idx[alphabet.index(S[i])].append(i + n)

    res = 0
    pos = 0
    for i in range(m):
        c = alphabet.index(T[i])
        if len(idx[c]) == 0:
            print(-1)
            break
        pos = idx[c][bisect.bisect_left(idx[c], pos)] + 1
        if pos >= n:
            pos -= n
            res += n
    else:
        res += pos
        print(res)


if __name__ == '__main__':
    resolve()