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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int main(){
    string s;
    cin>>s;
    string d;
    cin>>d;
    map<char,ll> m;
    ll cnt = 0;
     map<char,ll> p;
    for(unsigned ll i = 0;i<s.size();i++){
        cnt++;
           if(m[s[i]] != 1)
{
   p[s[i]] = i;
}     
        m[s[i]] = 1;
     
    }
    ll ans = 0;
    if(m[d[0]] == 1){
        ans += p[d[0]] +1;
    }
    else{
        cout<<"-1"<<"\n";
        return 0;
    }
  
    for(unsigned ll i = 1;i<d.size();i++){
if(m[d[i]] != 1){
    cout<<"-1"<<"\n";
    return 0;
}
else{
    if(p[d[i]] >p[d[i-1]]){
        ans += (p[d[i]] - p[d[i-1]]);
    }
    else{
        ans += (cnt - p[d[i-1]]);
        ans += (p[d[i]] );
    }
}
//cout<<ans<<"\n";
    }
    cout<<ans<<"\n";
}