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: Wrong Answer)

use proconio::{fastout, input};
use std::cmp::Ordering;
use std::collections::BinaryHeap;
use std::collections::HashSet;

// fn optmax<T:Ord + Copy>(x: Option<T>, v: T) -> Option<T> {
//     return Some(std::cmp::max(x.unwrap_or(v), v));
// }

#[derive(Clone, Eq, PartialEq)]
struct Node {
    s: String,
    is_left: bool,
    cost: i64 
}

impl Ord for Node {
    fn cmp(&self, other: &Node) -> Ordering {
        other.cost.cmp(&self.cost)
    }
}
impl PartialOrd for Node {
    fn partial_cmp(&self, other: &Node) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

fn is_palindrome(lhs: &String, rhs: &String) -> bool {
    let (ln, rn) = (lhs.len(), rhs.len());
    let n = std::cmp::min(ln, rn);

    for i in 0..n {
        if lhs.as_bytes()[i] != rhs.as_bytes()[rn - i - 1] { return false; }
    }
    return true;
}

#[fastout]
#[allow(non_snake_case)]
fn main() {
    input! {
        N: usize,
        Item: [(String, i64); N],
    }

    let mut heap = BinaryHeap::new();

    for item in &Item {
        heap.push(Node { s: item.0.clone(), is_left: true, cost: item.1 });
    }

    let mut lhs = HashSet::new();
    let mut rhs = HashSet::new();

    let mut res = None;
    
    while let Some(Node { s, is_left, cost}) = heap.pop() {
        let cache = if is_left { &mut lhs } else { &mut rhs };

        if cache.contains(&s) { continue; }
        cache.insert(s.clone());

        // コスト最小順で見つかった最初のコストを採用して終了
        if is_palindrome(&s, &s) { res = Some(cost); break; }

        for item in &Item {
            let ls = if is_left { &s } else { &item.0 };
            let rs = if is_left { &item.0 } else { &s };
            // println!("ls={}, rs={}, is_palindrome={}", ls, rs, is_palindrome(ls, rs));
            if !is_palindrome(&ls, &rs) { continue; }

            let ln = ls.len();
            let rn = rs.len();

            let ss = (if ln < rn { &rs[0..(rn - ln)] }
                              else       { &ls[(ln - rn + 1)..ln] }).to_string();

            heap.push(Node { s: ss, is_left: ln >= rn, cost: cost + item.1 });
        }
    }

    println!("{:?}", res.unwrap_or(-1));
}