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<iostream>
#include<stdio.h>
//#include <bits/stdc++.h>
#include<vector>
#include<float.h>
#include<iomanip>
#include<algorithm>
#include<string>
#include<cstring>
#include<math.h>
#include<cmath>
#include<sstream>
#include<set>
#include<map>
#include<queue>
#include <cassert>
#include <cmath>
#include<cstdint>

#define INF 1e9
#define rep(i,n)for(int i=0;(i)<(int)(n);i++)
#define REP(i,a,b)for(int i=(int)(a);(i)<=(int)(b);i++)
#define VEC(type, c, n) std::vector<type> c(n);for(auto& i:c)std::cin>>i;
#define vec(type,n) vector<type>(n)
#define vvec(m,n) vector<vector<int>> (int(m),vector<int>(n))
#define ALL(a)  (a).begin(),(a).end()

using namespace std;
using ll = long long;
using Graph = vector<vector<int>>;
using P = pair<ll, ll>;

bool seen[300000];
Graph G(300000);
bool f;
int g;
void dfs(int v, int cnt){
    seen[v] = true;
    if(v == g){
        f = true;
        return;
    }
    if(cnt == 2)return;
    for(int nv : G[v]){
        if(seen[nv])continue;
        dfs(nv, cnt+1);
    }

}

int main(){
    string s, t;
    cin >> s >> t;
    vector<vector<int>> v(26);                           //s[i]の位置を格納
    vector<bool> exist(26, false);               //sがその文字を持ってるかどうか
    rep(i, s.size()){
        v[s[i] - 'a'].push_back(i);
        exist[s[i] - 'a'] = true;
    }
    
    vector<ll>mem;                 //memはtの順番、cntはその文字の何番目か
    vector<ll>cnt(26, 0);
    ll cnt_ch = 0;
    mem.push_back(-1);

    rep(i, t.size()){

        if(!exist[t[i] - 'a']){
            cout << -1;
            return 0;
        }

        if(mem.back() >= v[t[i] - 'a'][cnt[t[i] - 'a']]){      //最後の文字が新しく追加される文字より大きかったらcnt_chを一つ増やして、cntを初期化
            rep(i, 26)cnt[i] = 0;
            cnt_ch++;
        }
        mem.push_back(v[t[i] - 'a'][cnt[t[i] - 'a']]);

        if(cnt[t[i] - 'a'] + 1 < v[t[i] - 'a'].size())cnt[t[i] - 'a']++;    //cntはv.size()を超えない
    }

    ll res = cnt_ch * s.size() + mem.back() + 1;
    cout << res << endl;
}