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: Time Limit Exceeded)

import bisect

s = list(input())
t = list(input())

# 辞書を作る
dictionary = {}

tt = t[:]
ss = dict(zip(list(range(len(s * 2))), s * 2))

while len(tt):
    _t = tt.pop(0)
    if _t in dictionary:
        continue
    indexes = [k for k, v in ss.items() if v == _t]
    if not len(indexes) > 0:
        print(-1)
        exit(0)
    for i in indexes:
        del ss[i]
    dictionary[_t] = indexes

result = 0
before = 0
sLen = len(s)
while len(t) > 0:
    _t = t.pop(0)
    next = dictionary[_t][bisect.bisect_left(dictionary[_t], before)] + 1
    result += next - before
    before = next
    if next >= sLen:
        before -= sLen

print(result)