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 FOR(i,a,b) for (int i = a; i < b; i++)
#define REP(i,b) FOR(i,0,b)
#define RFOR(i,a,b) for (int i = a-1; i >= b; i--)
#define RREP(i,a) RFOR(i,a,0)
#define REPALL(i,v) for (int i = 0; i < v.size(); i++)
#define RREPALL(i,v) for (int i = v.size()-1; i >= 0; i--)
#define ALL(v) v.begin(), v.end()
#define SORT(v) sort(v.begin(), v.end())
#define MIN_ELEMENT(v) min_element(v.begin(), v.end())
#define MAX_ELEMENT(v) max_element(v.begin(), v.end())
#define COUNT(v,n) count(v.begin(), v.end(), n)
void YES(bool flag) {cout<<(flag ? "YES" : "NO")<<endl;}
void Yes(bool flag) {cout<<(flag ? "Yes" : "No")<<endl;}
void yes(bool flag) {cout<<(flag ? "yes" : "no")<<endl;}
typedef long long ll;
typedef unsigned long long ull;
const int INF = 1e7;
const ll MOD = 1e9 + 7;

string s, t;
vector<vector<int>> appr(26);

bool check()
{
    vector<bool> exist(26, false);
    REPALL(i,s) {
        exist[s[i] - 'a'] = true;
        appr[s[i] - 'a'].push_back(i);
        appr[s[i] - 'a'].push_back(i + s.size());
    }
    REP(i,26)
        SORT(appr[i]);
    bool rtn = true;
    REPALL(i,t)
        rtn &= exist[t[i] - 'a'];
    return rtn;
}

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

    if (!check()) {
        cout<<-1<<endl;
        return 0;
    }

    ll idx_s = 0ll, idx_t = 0ll;
    while (idx_t < t.size()) {
        idx_s += *lower_bound(ALL(appr[t[idx_t]-'a']), idx_s % s.size())
                - idx_s % s.size();
        idx_t++;
    }
    cout<<idx_s+1<<endl;

    return 0;
}