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: Accepted)

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <fstream>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <vector>

#define len(X) ((int)(X).size())

#ifdef __LOCAL
#define DBG(X) cout << #X << "=" << (X) << '\n';
#define SAY(X) cout << (X) << '\n';
#else
#define DBG(X)
#define SAY(X)
#endif

#define M_PI       3.14159265358979323846

using namespace std;

using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

const int INT_INF = (int)(2e9);
const ll  LL_INF = (ll)(2e18);

const int NIL = -1;
static mt19937 _g(time(nullptr));

inline ll randint(ll a, ll b) { ll w = (_g() << 31LL) ^ _g(); return a + w % (b - a + 1); }
inline void fast_io() { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); };
template<typename T> inline T sign(T x) { return T(x > 0) - T(x < 0); }
template<typename T, typename S> inline ostream& operator<<(ostream& os, const pair<T, S> p) { cout << "[" << p.first << ";" << p.second << "]"; return os; }
template<typename T> inline ostream& operator<<(ostream& os, const vector<T>& v) { for (auto el : v) cout << el << " "; return os; }
template<typename T> inline T fetch() { T ret; cin >> ret; return ret; }
template<typename T> inline vector<T> fetch_vec(int sz) { vector<T> ret(sz); for (auto& elem : ret) cin >> elem; return ret; }

ll solve()
{
    string s, t;

    cin >> s >> t;

    unordered_map<char, vector<int>> charPos;

    for (int i = 0; i < s.length(); ++i)
        charPos[s[i]].push_back(i);
    
    ll counter = 0;
    ll lastPos = -1;

    for (int i = 0; i < t.length(); ++i)
    {
        if (charPos[t[i]].empty())
            return -1;

        auto it = std::upper_bound(charPos[t[i]].begin(), charPos[t[i]].end(), lastPos);

        if (it == charPos[t[i]].end())
        {
            ++counter;
            lastPos = -1;

            it = std::upper_bound(charPos[t[i]].begin(), charPos[t[i]].end(), lastPos);
        }

        lastPos = *it;
    }

    return counter * s.length() + (lastPos + 1);
}

int main()
{
    cout << solve() << endl;

    return 0;
}