Problem:
Score : 600 points

Problem Statement

We have N strings of lowercase English letters: S_1, S_2, \cdots, S_N.

Takahashi wants to make a string that is a palindrome by choosing one or more of these strings - the same string can be chosen more than once - and concatenating them in some order of his choice.

The cost of using the string S_i once is C_i, and the cost of using it multiple times is C_i multiplied by that number of times.

Find the minimum total cost needed to choose strings so that Takahashi can make a palindrome.

If there is no choice of strings in which he can make a palindrome, print -1.

Constraints

1 \leq N \leq 50

1 \leq |S_i| \leq 20

S_i consists of lowercase English letters.

1 \leq C_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
S_1 C_1
S_2 C_2
:
S_N C_N

Output

Print the minimum total cost needed to choose strings so that Takahashi can make a palindrome, or -1 if there is no such choice.

Sample Input 1

3
ba 3
abc 4
cbaa 5

Sample Output 1

7

We have ba, abc, and cbaa.

For example, we can use ba once and abc once for a cost of 7, then concatenate them in the order abc, ba to make a palindrome.
Also, we can use abc once and cbaa once for a cost of 9, then concatenate them in the order cbaa, abc to make a palindrome.

We cannot make a palindrome for a cost less than 7, so we should print 7.

Sample Input 2

2
abcab 5
cba 3

Sample Output 2

11

We can choose abcab once and cba twice, then concatenate them in the order abcab, cba, cba to make a palindrome for a cost of 11.

Sample Input 3

4
ab 5
cba 3
a 12
ab 10

Sample Output 3

8

We can choose a once, which is already a palindrome, but it is cheaper to concatenate ab and cba.

Sample Input 4

2
abc 1
ab 2

Sample Output 4

-1

We cannot make a palindrome, so we should print -1.

Solution in Rust (1.42.0) (judge verdict: Accepted)

use std::collections::HashMap;

fn main() {
    proconio::input! {
        n: usize,
        sc: [(String, u64); n],
    }

    let rev: Vec<_> = sc
        .iter()
        .map(|(s, _)| s.chars().rev().collect::<String>())
        .collect();

    let mut cost = u64::max_value();
    for i in 0..n {
        let mut solver = Solver::new(n, &sc, &rev);
        cost = cost.min(if sc[i].0 == solver.rev[i] {
            sc[i].1
        } else {
            solver.calc(sc[i].0.as_str(), sc[i].1, true)
        });
    }

    if cost < u64::max_value() {
        println!("{}", cost);
    } else {
        println!("-1");
    }
}

struct Solver<'a> {
    done: HashMap<(&'a str, bool), u64>,
    n: usize,
    sc: &'a Vec<(String, u64)>,
    rev: &'a Vec<String>,
}

impl<'a> Solver<'a> {
    fn new(n: usize, sc: &'a Vec<(String, u64)>, rev: &'a Vec<String>) -> Solver<'a> {
        Solver {
            done: HashMap::new(),
            n: n,
            sc: sc,
            rev: rev,
        }
    }

    fn calc(&mut self, s: &'a str, cost: u64, is_left: bool) -> u64 {
        let mut v = u64::max_value();
        if if let Some(&c) = self.done.get(&(s, is_left)) { cost >= c } else { false } {
            v
        } else {
            self.done.insert((s, is_left), cost);
            for i in 0..self.n {
                let oppo = if is_left { &self.rev[i] } else { &self.sc[i].0 };
                let next = if s.starts_with(oppo) {
                    Some((&s[oppo.len()..], is_left))
                } else if oppo.starts_with(&s) {
                    Some((&oppo[s.len()..], !is_left))
                } else {
                    None
                };
                if let Some((next, is_left)) = next {
                    let cost = cost + self.sc[i].1;
                    v = v.min(if next == next.chars().rev().collect::<String>() {
                        cost
                    } else {
                        self.calc(next, cost, is_left)
                    });
                }
            }
            v
        }
    }
}