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

ds = {}  # ｓに含まれる文字それぞれの数
dsl = {}  # ｓの文字の位置
for i in range(len(s)):
    if s[i] not in ds.keys():
        ds[s[i]] = 0
        dsl[s[i]] = []
    ds[s[i]] += 1
    dsl[s[i]].append(i)

for i in range(len(t)):
    if t[i] not in ds.keys():
        ds[t[i]] = 0

#print(ds, dsl)

ans = 0
now = -1#ここが-1じゃないと、最初の遷移でsの最初の文字を使えない
for i in range(len(t)):
    if ds[t[i]] == 0:  # ｓにｔを構成する文字がない場合
        print("-1")
        exit()
    elif ds[t[i]] == 1:
        if dsl[t[i]][0] > now:  # ｔの次の文字が現在地より右にある場合
            ans += (dsl[t[i]][0] - now)
        else:  # ｔの次の文字が現在地より左にある場合
            ans += dsl[t[i]][0] + (len(s) - now)
        now = dsl[t[i]][0]
    else:
        index = bisect.bisect_right(dsl[t[i]], now)
        if index == len(dsl[t[i]]):
            index = -1
        if dsl[t[i]][index] > now:  # ｔの次の文字が現在地より右にある場合
            ans += (dsl[t[i]][index] - now)
            now = dsl[t[i]][index]
        else:  # ｔの次の文字が現在地より左にある場合
            ans += dsl[t[i]][0] + (len(s) - now)
            now = dsl[t[i]][0]

print(ans)