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 itertools::{iproduct, Itertools};
use proconio::{input, marker::Bytes};
use std::cmp::Reverse;
use std::collections::{BinaryHeap, HashMap};

fn main() {
    input!(s: [(Bytes, u64)]);

    type Rest = (Vec<u8>, bool);

    let mut dist = HashMap::<Rest, u64>::new();
    let mut que = BinaryHeap::<(Reverse<u64>, Rest)>::new();

    for ((s, c), &left) in iproduct!(&s, &[true, false]) {
        let c = *c;
        dist.entry((s.clone(), left))
            .and_modify(|d| *d = c.min(*d))
            .or_insert(c);
        que.push((Reverse(c), (s.clone(), left)))
    }

    while let Some((Reverse(dv), (v, left))) = que.pop() {
        let adj = s.iter().flat_map(|(s, c)| {
            if left {
                let w = &v.iter().copied().rev().collect_vec();
                if s.ends_with(w) {
                    Some((&s[..s.len() - v.len()], false))
                } else if w.ends_with(s) {
                    Some((&v[s.len()..], true))
                } else {
                    None
                }
            } else {
                let w = &v.iter().copied().rev().collect_vec();
                if s.starts_with(w) {
                    Some((&s[v.len()..], true))
                } else if w.starts_with(s) {
                    Some((&v[..v.len() - s.len()], false))
                } else {
                    None
                }
            }
            .map(|(s, left)| ((s.into(), left), c))
        });

        for (w, &c) in adj {
            let alt = dv + c;
            if let Some(dw) = dist.get_mut(&w) {
                if *dw > alt {
                    *dw = alt;
                    que.push((Reverse(alt), w.clone()));
                }
            } else {
                dist.insert(w.clone(), alt);
                que.push((Reverse(alt), w.clone()));
            }
        }
    }

    let ans = dist
        .iter()
        .filter(|&((s, _), _)| s.iter().eq(s.iter().rev()))
        .map(|(_, d)| d)
        .min();
    if let Some(ans) = ans {
        println!("{}", ans)
    } else {
        println!("-1")
    }
}