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: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> P;
typedef vector<ll> Vec;

#define REP(i, a, b) for(ll i=(a); i<(b); i++)
#define rep(i, n) REP(i, 0, n)
const int INF=1e9;
const int MOD =  1000000007;
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl;
#define ALL(v) v.begin(), v.end()
#define dbg(x_) cerr << #x_ << ":" << x_ << endl;
#define pb(x) push_back(x)
#define REPM(i, mp) for (auto i = mp.begin(); i != mp.end(); ++i)
#define dbgmap(mp) for (auto i = mp.begin(); i != mp.end(); ++i) { cout << i->first <<":"<<i->second << endl;}
#define sum(v) accumulate(ALL(v),0)
#define lb(v, c) lower_bound((v).begin(), (v).end(), c) - (v).begin()
#define ub(v, c) upper_bound((v).begin(), (v).end(), c) - (v).begin()

template<typename T1, typename T2>
ostream &operator<<(ostream &s, const pair<T1, T2> &p) { return s<<"("<<p.first<<", "<<p.second<<")"; }

// vector
template<typename T>
ostream &operator<<(ostream &s, const vector<T> &v) {
    int len=v.size();
    for(int i=0; i<len; ++i) {
        s<<v[i];
        if(i<len-1) s<<"	";
    }
    return s;
}

// 2 dimentional vector
template<typename T>
ostream &operator<<(ostream &s, const vector<vector<T> > &vv) {
    int len=vv.size();
    for(int i=0; i<len; ++i) {
        s<<vv[i]<<endl;
    }
    return s;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    string s, t;
    cin>>s>>t;
    ll tlength=t.size();
    ll slength=s.size();
    vector<vector<ll>> smp(30);
    rep(i, slength) {
        smp[s[i]-'a'].pb(i+1);
    }

    ll ans=0;
    ll now=0;
    ll loc=0;
    rep(i, tlength) {
        vector<ll> &v = smp[t[i]-'a'];
        if(v.size()<=0){
            cout << -1 << endl;
            return 0;
        }else{
            auto itr = upper_bound(ALL(v), now);
            if(itr==v.end()){
                loc = *v.begin();
                ans += slength;
            }else{
                loc = *itr;

            }
            //dbg(loc);
            now=loc;
        }

    }
    ans += loc;
    cout<<ans<<endl;
    return 0;
}