We will be greedy in constructing the string.

1. Initially we have matches 0 characters

2. We will then look for the earliest location of T[0]

3. Then, we will look for the earliest location in [0, n] in S for T[0]. Let this position be p.

4. Then, we will look for the earliest location in [p, n] in S for T[1]

5. If T[1] is not present in [p, n] but is in [0, p - 1], we will go back and start another subsequence.

-----

After that the length of the string we need is = (No_of_Concatenations)|S| + Matched_Prefix

-----

int main()

{

string S, T;

cin >> S >> T;

const int NO_OF_ALPHABETS = 26;

vector <vector <int> > locations(NO_OF_ALPHABETS);

for(int i = 0; i < S.size(); i++)

{

locations[S[i] - 'a'].push_back(i);

}

int no_of_concatenations = 1, matched_prefix = -1;

for(int i = 0; i < T.size(); i++)

{

if(locations[T[i] - 'a'].size() == 0)

{

cout << "-1\n";

return 0;

}

auto it = upper_bound(locations[T[i] - 'a'].begin(), locations[T[i] - 'a'].end(), matched_prefix);

if(it == locations[T[i] - 'a'].end())

{

no_of_concatenations++;

matched_prefix = locations[T[i] - 'a'][0];

}

else

{

matched_prefix = *it;

}

}

long long final_length = (no_of_concatenations - 1)*1LL*S.size() + (matched_prefix + 1);

cout << final_length << "\n";

return 0;

}