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)

#[allow(unused_imports)]
use itertools::Itertools;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

#[derive(Clone, Ord, PartialOrd, Eq, PartialEq, Debug)]
enum D {
    L,
    R,
}

fn f(l: &Vec<char>, r: &Vec<char>) -> Option<(D, Vec<char>)> {
    if !r.iter().rev().zip(l).all(|(c1, c2)| c1 == c2) {
        return None;
    }
    if l.len() == r.len() {
        return Some((D::L, vec![]));
    }
    if l.len() > r.len() {
        Some((D::L, Vec::from(&l[r.len()..])))
    } else {
        Some((D::R, Vec::from(&r[..r.len() - l.len()])))
    }
}

fn g(v: &Vec<char>) -> bool {
    v.iter().rev().eq(v)
}

fn main() {
    input! {
        n: usize,
        mut sc: [(Chars, i64); n],
    }

    use std::cmp::{max, Reverse};
    let mut res = None;
    let mut memo = std::collections::BTreeMap::new();
    let mut q = std::collections::BinaryHeap::new();
    for (s, c) in &sc {
        q.push((Reverse(*c), (D::L, s.clone())));
    }
    sc.push((vec![], 0));
    while let Some((Reverse(c), s)) = q.pop() {
        if *memo.entry(s.clone()).or_insert(std::i64::MAX) < c {
            continue;
        }
        for j in 0..n + 1 {
            let x = if s.0 == D::L {
                f(&s.1, &sc[j].0)
            } else {
                f(&sc[j].0, &s.1)
            };
            if let Some(s) = x {
                if g(&s.1) {
                    res = max(res, Some(Reverse(c + sc[j].1)));
                    continue;
                }
                let r = memo.entry(s.clone()).or_insert(std::i64::MAX);
                let c = c + sc[j].1;
                if *r <= c {
                    continue;
                }
                *r = std::cmp::min(*r, c);
                q.push((Reverse(c), s));
            }
        }
    }

    println!("{}", res.unwrap_or(Reverse(-1)).0);
}