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<algorithm>//sort,二分探索,など
#include<bitset>//固定長bit集合
#include<cmath>//pow,logなど
#include<complex>//複素数
#include<deque>//両端アクセスのキュー
#include<functional>//sortのgreater
#include<iomanip>//setprecision(浮動小数点の出力の誤差)
#include<iostream>//入出力
#include<map>//map(辞書)
#include<numeric>//iota(整数列の生成),gcdとlcm(c++17)
#include<queue>//キュー
#include<set>//集合
#include<stack>//スタック
#include<string>//文字列
#include<utility>//pair
#include<list>
#include<vector>//可変長配列
using namespace std;
#define rep(i,a) for(int i=0;i<a;i++)
#define mp make_pair
#define pb push_back
#define ll __int64
#define __int64 long long
#define P pair<int,int>
#define PP pair<P,int>
#define F first
#define S second
int dx[8] = { 1,0,-1,0,1,-1,-1,1 };
int dy[8] = { 0,1,0,-1,1,1,-1,-1 };


string s, t;
int rui[26][222222];

int q(int c, int l, int r) {
    if (l == 0)
        return rui[c][r];
    return rui[c][r] - rui[c][l - 1];
}

int main() {
    cin >> s >> t;
    s = s + s;

    int ss = s.size();

    rep(i, ss) {
        rui[s[i] - 'a'][i]++;
    }
    rep(i, 26) {
        rep(j, ss) {
            rui[i][j + 1] += rui[i][j];
        }
    }


    rep(i,t.size()) {
        if (q(t[i] - 'a', 0, ss) == 0) {
            cout << -1 << endl;
            return 0;
        }
    }
 
        
    int ans = 0;
    int now = 0;
    rep(i, t.size()) {
        int ng = now-1, ok = ss;
        while (abs(ok - ng) > 1) {
            int mid = (ng + ok) / 2;
            int res = q(t[i]-'a', now, mid);
            if (res >= 1)
                ok = mid;
            else
                ng = mid;
        }
        //cout << t[i] << " " << ok << endl;
        if (ok == ss) {
            ans += ss;
            now = 0;
        }
        else {
            now = ok + 1;
        }
    }

    cout << ans+now << endl;


    return 0;
}