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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
 
#define rep(i,n) for(ll i=0;i<n;++i)
#define all_map(pos,mp) for(ll pos=mp.begin();pos!=mp.end();pos -++)
#define ALL(a) (a).begin(),(a).end()

using namespace std;
using ll  = long long;
using pll = pair<ll, ll>;

const ll MOD  = 1e9 + 7;
const ll LINF = 1LL << 62;
const int INF = 1e9 + 7;

int main(){
    string s, t;
    cin >> s >> t;

    vector<vector<ll>> vec(26);
    rep(i, s.size()){
        vec[s[i]-'a'].push_back(i+1);
    }

    ll ans = 0;
    ll pos = -1;
    for(char c : t){
        if(vec[c-'a'].empty()){
            cout << -1 << endl;
            return 0;
        }

        auto itr = upper_bound(ALL(vec[c-'a']), pos);
        if(itr == vec[c-'a'].end()){
            ans += s.size();
            pos = vec[c-'a'].front();
        }
        else pos = *itr;
    }
    
    cout << ans + pos;
}