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 Ruby (2.3.3) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int,int> P;

#define p_ary(ary,a,b,i) do { cout << "["; for (int (i) = (a);(i) < (b);++(i)) cout << ary[(i)] << ((b)-1 == (i) ? "" : ", "); cout << "]\n"; } while(0)
#define p_map(map,it) do {cout << "{";for (auto (it) = map.begin();;++(it)) {if ((it) == map.end()) {cout << "}\n";break;}else cout << "" << (it)->first << "=>" << (it)->second << ", ";}}while(0)

int main() {
    string s,t;
    cin >> s >> t;
    int n = s.size(),m = t.size();
    vector<int> x[26];
    for (int i = 0;i < n;++i) x[s[i]-'a'].push_back(i);
    int ans = 0,p = 0;
    for (int i = 0;i < m;++i) {
        int c = t[i]-'a';
        if (x[c].empty()) {
            cout << "-1" << endl;
            return 0;
        }
        auto it = lower_bound(x[c].begin(),x[c].end(),p);
        if (it == x[c].end()) {
            ans++;
            p = x[c][0];
        } else {
            p = *it;
        }
    }
    cout << (ll)ans*n+p+1 << endl;
    return 0;
}