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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.PrintWriter;
import java.util.*;

public class Main {

    private static Scanner in = new Scanner(System.in);
    private static PrintWriter out = new PrintWriter(System.out);

    private static final long MOD = (long) 1e9 + 7;

    public static void main(String[] args) {
        solve();
        out.flush();
    }

    private static void solve() {
        String s = in.next();
        String t = in.next();
        int sn = s.length();
        int tn = t.length();

        Map<Character, List<Integer>> sMap = new HashMap<>();
        char[] chars = s.toCharArray();
        for (int i = 0; i < chars.length; i++) {
            char ch = chars[i];
            if (sMap.containsKey(ch)) {
                sMap.get(ch).add(i);
            } else {
                List<Integer> l = new ArrayList<>();
                l.add(i);
                sMap.put(ch, l);
            }
        }

        Map<Character, int[]> map = new HashMap<>();
        sMap.forEach((k, v) -> {
            v.sort((Comparator.comparingInt(o -> o)));
            map.put(k, v.stream().mapToInt(i->i).toArray());
        });

        // out.println(sMap);
        long ans = 0;
        int before = -1;
        chars = t.toCharArray();
        for (char ch : chars) {
            int[] possible = map.get(ch);
            if (possible == null || possible.length == 0) {
                ans = -1;
                break;
            }

            int minIndex = nextIndex(possible, before, sn);
            int min = distance(minIndex, before, sn);
            ans += min;
            before = minIndex;
            // out.println("Char " + ch + " minIndex " + minIndex + " min " + min);
        }

        out.println(ans);
    }

    private static int nextIndex(int[] possible, int before, int sn) {
        int pt = Arrays.binarySearch(possible, before);
        // out.println("Binary search " + Arrays.toString(possible) + " before " + before + " result " + pt + " bit not " + ~pt);
        if (pt >= 0)
             return possible[(pt + 1) % possible.length];
        pt = ~pt;

        return possible[pt % possible.length];
    }

    private static int distance(int i, int before, int sn) {
        if (i == before)
            return sn;
        if (i > before)
            return i - before;
        return i + sn - before;
    }

}