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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::Read;
use std::str::SplitWhitespace;
use std::str::FromStr;
trait BinarySearch<T> {
    fn front_position(&self, x: T) -> usize;
    fn back_position(&self, x: T) -> usize;
    fn bs_position(v: &[T], x: T, is_back: bool) -> usize;
}

impl<T> BinarySearch<T> for Vec<T>
where
    T: Ord,
{
    fn front_position(&self, x: T) -> usize {
        Self::bs_position(self, x, true)
    }

    fn back_position(&self, x: T) -> usize {
        Self::bs_position(self, x, false)
    }

    fn bs_position(v: &[T], x: T, is_front: bool) -> usize {
        let len = v.len();
        if len == 1 {
            return if x < v[0] || (is_front && x == v[0]) {
                0
            } else {
                1
            };
        }
        let i = len / 2;
        if x < v[i] || (is_front && x == v[i]) {
            Self::bs_position(&v[..i], x, is_front)
        } else {
            Self::bs_position(&v[i..], x, is_front) + i
        }
    }
}

pub fn read_stdin() -> String {
    let mut s = String::new();
    ::std::io::stdin().read_to_string(&mut s).unwrap();
    s
}

pub trait Parse<T> {
    fn parse(&mut self) -> T;
}

impl<'a, T> Parse<T> for SplitWhitespace<'a>
where
    T: FromStr,
{
    fn parse(&mut self) -> T {
        let word = self.next().unwrap();
        word.parse().ok().unwrap()
    }
}

fn main() {
    let source = read_stdin();
    let mut iter = source.split_whitespace();
    let s_: String = iter.parse();
    let s = s_.chars().collect::<Vec<_>>();
    let t_: String = iter.parse();
    let t = t_.chars().collect::<Vec<_>>();

    let mut m = std::collections::HashMap::new();
    let alphabet = "abcdefghijklmnopqrstuvwxyz";
    for c in alphabet.chars() {
        m.insert(c, Vec::new());
    }

    let s_len = s.len();
    for (i, c) in s.into_iter().enumerate() {
        if let Some(v) = m.get_mut(&c) {
            v.push(i);
        }
    }

    let mut ans = 0;
    let mut curr = 0;
    for c in t {
        if let Some(v) = m.get(&c) {
            if v.is_empty() {
                println!("-1");
                return;
            }
            let mut pos = v.front_position(curr);
            if pos == v.len() {
                ans += s_len - curr;
                curr = 0;
                pos = v.front_position(curr);
            }
            ans += v[pos] + 1 - curr;
            curr = v[pos] + 1;
        }
    }

    println!("{}", ans);
}