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 C# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Collections;

namespace ConsoleApplication2
{
    class StringWithSerchIndex
    {
        string moji;
        int len;
        Dictionary<char, List<int>> index;
        public StringWithSerchIndex(string v)
        {
            len = v.Length;
            moji = v;
        }
        public void makeIndex()
        {
            index = new Dictionary<char, List<int>>();
            for (int i = 0; i < len; i++)
            {
                if (!index.ContainsKey(moji[i]))
                {
                    index.Add(moji[i], new List<int>());
                }
                index[moji[i]].Add(i);
            }
        }

        public int Length
        {
            get { return len; }
        }
        public char this[int i]
        {
            get { return moji[i]; }
        }

        public int IndexOf(char tar)
        {
            if (index[tar] == null) return -1;

            return index[tar][0];
        }
        public int IndexOf(char tar, int startIndex)
        {
            if (index[tar] == null) return -1;
            foreach(int ret in index[tar])
            {
                if (ret >= startIndex) return ret;
            }
            return -1;
        }
    }
    class Program
    {

        static void Main(string[] args)
        {
          
            string ss = Console.ReadLine();
            string t = Console.ReadLine();
            StringWithSerchIndex s = new StringWithSerchIndex(ss);

            s.makeIndex();

            int sl = s.Length;
            int n = t.Length;

            int[] find = new int[n];

            long ans = 0;

            find[0] = s.IndexOf(t[0]);
            ans = find[0];
            if (find[0] < 0)
            {
                ans = -1;
                Console.WriteLine(ans);
                return;
            }


            for (int i = 1; i < n; i++)
            {
                int f1 = s.IndexOf(t[i]);//最速
                if (f1 < 0)
                {
                    ans = -1;
                    Console.WriteLine(ans);
                    return;
                }
                int f2 = s.IndexOf(t[i], find[i - 1] + 1);//前の文字より後ろで最速
                //前の文字より後ろで見つかった場合
                if (f2 > -1)
                {
                    find[i] = f2;
                }
                else
                {
                    find[i] = f1;
                }
                if (find[i] > find[i - 1])
                {
                    ans += (find[i] - find[i - 1]);
                }
                else
                {
                    ans += sl - find[i - 1] + find[i];
                }
            }
            Console.WriteLine(ans+1);

        }
        

    }
}