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(clippy::needless_range_loop)]
#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

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

type Remainder = (Vec<u8>, bool); // Remaining string and bool whether the string is on the left side

fn is_parindrome(bytes: &[u8]) -> bool {
    bytes.iter().eq(bytes.iter().rev())
}

fn main() {
    input! {
        sc: [(Bytes, usize)]
    }

    let mut costs = HashMap::<Remainder, usize>::new();
    let mut queue = BinaryHeap::<(Reverse<usize>, Remainder)>::new();
    for ((s, c), &on_left) in iproduct!(&sc, &[false, true]) {
        let cost = *c;
        costs
            .entry((s.clone(), on_left))
            .and_modify(|cost_prev| *cost_prev = cost.min(*cost_prev))
            .or_insert(cost);
        queue.push((Reverse(cost), (s.clone(), on_left)));
    }

    // Dijkstra on Remainder
    while let Some((Reverse(cost), (orig, on_left))) = queue.pop() {
        let reversed: Vec<_> = orig.iter().rev().copied().collect();
        sc.iter()
            .flat_map(|(dest, cost)| {
                let on_right = !on_left;
                if on_left {
                    if reversed.ends_with(&dest) {
                        return Some((&orig[dest.len()..], true, cost));
                    }
                    if dest.ends_with(&reversed) {
                        return Some((&dest[..dest.len() - orig.len()], false, cost));
                    }
                }
                if on_right {
                    if reversed.starts_with(&dest) {
                        return Some((&orig[..orig.len() - dest.len()], false, cost));
                    }
                    if dest.starts_with(&reversed) {
                        return Some((&dest[orig.len()..], true, cost));
                    }
                }
                None
            })
            .for_each(|(dest, on_left, cost_diff)| {
                let remainder = (dest.into(), on_left);
                let cost_sum = cost + cost_diff;
                if let Some(mut_v) = costs.get_mut(&remainder) {
                    if *mut_v > cost_sum {
                        *mut_v = cost_sum;
                        queue.push((Reverse(cost_sum), remainder.clone()))
                    }
                } else {
                    costs.insert(remainder.clone(), cost_sum);
                    queue.push((Reverse(cost_sum), remainder.clone()))
                }
            });
    }

    let ans = costs
        .iter()
        .filter(|&((remaining, _), _)| is_parindrome(remaining))
        .map(|(_, &cost)| cost as isize)
        .min()
        .unwrap_or(-1);

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