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;
typedef long long ll;
#define rep(i, a, b) for (int i = a; i < (int)b; i++)
#define all(v) v.begin(), v.end()
const int MOD = 1e9 + 7;
const int INF = 1e9;
//'A' = 65, 'Z' = 90, 'a' = 97, 'z' = 122

int main() {
    string s, t;
    cin >> s >> t;
    vector<vector<int>> v(26);
    int len = s.size();
    rep(i, 0, s.size()) {
        v[s[i] - 'a'].push_back(i);
        v[s[i] - 'a'].push_back(i + len);
    }
    ll ans = 0;
    int p = 0;
    rep(i, 0, t.size()) {
        int c = t[i] - 'a';
        if ((int)v[c].size() == 0) {
            cout << -1 << endl;
            return 0;
        }
        p = *lower_bound(all(v[c]), p);
        if (p > len - 1) {
            ans += len;
            p -= len;
        }
        p++;
    }
    ans += p;
    cout << ans << endl;
    return 0;
}