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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

from bisect import *

S = input()
T = input()

ds = {}
for i,s in enumerate(S) :
    if s not in ds :
        ds[s] = [i]
    else :
        ds[s].append(i)
L = []
ind = -1
flg = 0
for t in T :
    if t not in ds :
        ans = -1
        flg = 1
        break
if flg == 0 :
    for t in T :
        if ind >= ds[t][-1] :
            ind = ds[t][0]
            L.append(ind)
        else :
            ind = ds[t][bisect_right(ds[t], ind)]
            L.append(ind)

    cnt = 0
    val = L[0]
    for i in range(1,len(L)) :
        if val >= L[i] :
            val = L[i]
            cnt += 1

    ans = cnt * len(S) + L[-1] + 1
print(ans)