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>
using namespace std;
#define rep(i,n) for(int i=0; i<n; i++)
#define REP(i,m,n) for(ll i=(ll)(m);i<(ll)(n);i++)
long long mo = 1e9 + 7;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> Pii;
typedef pair<ll,ll> Pll;
template<class T, class S> void cmin(T &a, const S &b) { if (a > b)a = b; }
template<class T, class S> void cmax(T &a, const S &b) { if (a < b)a = b; }
template<class A>void PR(A a,ll n){rep(i,n){if(i)cout<<' ';cout<<a[i];}cout << "\n";}
ld PI=3.14159265358979323846;

int main(){
    string S,T;
    cin >> S >> T;
    ll N = S.size();
    ll M = T.size();
    vector<queue<ll>> v(26);
    rep(i,N){
        v[S[i] - 'a'].push(i);
    }
    ll ans = -1;
    ll tmp = -1;
    rep(i,M){
        if(v[T[i] - 'a'].size() == 0){
            cout << -1 << endl;
            return 0;
        }
        tmp = v[T[i] - 'a'].front();
        v[T[i] - 'a'].pop();
        if(ans < tmp){
            ans = tmp;
            v[T[i] - 'a'].push(tmp);        
        }else{
            ll q = (ans-tmp)/N + 1;
            ans = tmp + q*N;
            v[T[i] - 'a'].push(tmp);        
        }
    }
    cout << ans+1 << endl;
}