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 proconio::input;

#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars, Usize1};
#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::ops::*;

#[derive(Clone, Debug, Default)]
struct Struct;

#[derive(Clone, Copy, Debug, Default, Hash, PartialEq, Eq)]
struct Remain<'a> {
    remstr: &'a [u8],
    is_reverse: bool,
}

#[derive(Clone, Copy, Debug, Default)]
struct RemainCost<'a> {
    cost: i64,
    remain: Remain<'a>,
}

impl<'a> Ord for RemainCost<'a> {
    fn cmp(&self, other: &Self) -> Ordering {
        self.cost.cmp(&other.cost).reverse()
    }
}
impl<'a> PartialOrd for RemainCost<'a> {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(self.cmp(&other))
    }
}
impl<'a> PartialEq for RemainCost<'a> {
    fn eq(&self, other: &Self) -> bool {
        self.cost == other.cost
    }
}
impl<'a> Eq for RemainCost<'a> {}

fn is_kaibun(s: &[u8]) -> bool {
    let len = s.len();
    for i in 0..len / 2 {
        if s[i] != s[len - 1 - i] {
            return false;
        }
    }
    true
}

//#[proconio::fastout]
fn main() {
    input! {
        n: usize,
        v: [(Bytes,i64);n],
    }
    let mut vr = Vec::with_capacity(n);
    for a in &v {
        let mut s = a.0.clone();
        s.reverse();
        vr.push((s, a.1));
    }

    let mut h: HashMap<Remain, i64> = HashMap::new(); // remain -> min_cost  ループ検出用
    let mut q: BinaryHeap<RemainCost> = BinaryHeap::new();
    for a in &v {
        q.push(RemainCost {
            cost: a.1,
            remain: Remain {
                remstr: &a.0[..],
                is_reverse: false,
            },
        });
    }

    const MAX_COST: i64 = 1000_000_000_000_000_000;
    let mut ans = MAX_COST;
    while let Some(s) = q.pop() {
        if s.cost >= ans {
            continue;
        }

        if let Some(val) = h.get_mut(&s.remain) {
            if *val <= s.cost {
                continue; // 以前にもっと低いコストで調べたなら、スキップする
            } else {
                *val = s.cost;
            }
        } else {
            h.insert(s.remain, s.cost);
        }
        let RemainCost { cost, remain } = s;
        let Remain { remstr, is_reverse } = remain;

        if is_kaibun(remstr) {
            ans = cost;
        } else {
            let vv = if is_reverse { &v } else { &vr };
            for (add_str, add_cost) in vv {
                let len = min(add_str.len(), remstr.len());
                if &remstr[0..len] == &add_str[0..len] {
                    // println!(
                    //     "cmp {:?} {:?}",
                    //     &remstr[0..len].to_vec(),
                    //     &add_str[0..len].to_vec()
                    // );
                    let rem = if add_str.len() < remstr.len() {
                        Remain {
                            remstr: &remstr[len..],
                            is_reverse: is_reverse,
                        }
                    } else {
                        Remain {
                            remstr: &add_str[len..],
                            is_reverse: !is_reverse,
                        }
                    };
                    // println!(
                    //     "push cost={}, remain={:?}, reverse={}",
                    //     cost + add_cost,
                    //     rem.remstr,
                    //     rem.is_reverse
                    // );
                    q.push(RemainCost {
                        cost: cost + add_cost,
                        remain: rem,
                    });
                }
            }
        }
    }

    println!("{}", if ans == MAX_COST { -1 } else { ans });
}