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 M_PI       3.14159265358979323846

using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); }
inline int readInt() { int x; scanf("%d", &x); return x; }

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<int, PII> TIII;
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;


//container util

//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a)*(a))
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())

//repetition
//------------------------------------------
#define FOR(i,s,n) for(int i=s;i<(int)n;++i)
#define REP(i,n) FOR(i,0,n)
#define MOD 1000000007

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)


struct Edge {
    int to, from;
    ll cost;
    Edge(int from, int to, ll cost): from(from), to(to), cost(cost) {}
};

typedef vector<vector<Edge>> AdjList;
AdjList graph;

int main() {
    //cin.tie(0);
    //ios::sync_with_stdio(false);
    //cout << fixed << setprecision(10);

    string s, t; cin >> s >> t;

    int n = s.size(), m = t.size();

    vector<vector<int>> v(26);
    
    REP(i, n){
        v[s[i]-'a'].push_back(i);
    }

    ll ans = 1;
    int idx=0;
    REP(i, m){
        int c = t[i]-'a';
        if(v[c].size()==0){
            cout << -1 << endl;
            return 0;
        }
        if(upper_bound(v[c].begin(), v[c].end(), idx) == v[c].end()){
            ans += n-idx-1;
            //cout << n-idx-1 << endl;
            ans += v[c][0]+1;
            //cout << v[c][0]+1 << endl;
            idx = v[c][0];
        }else{
            auto idx2 = upper_bound(v[c].begin(), v[c].end(), idx);

            //cout << idx << " " << *idx2 << endl;
                ans += *idx2-idx;
            
            idx = *idx2;
        }
        //cout << idx << endl;
    }
    cout << ans << endl;
    return 0;

}