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 Python (3.8.2) (judge verdict: Accepted)

from collections import defaultdict
from heapq import heappush, heappop, heapify
def main():
    N = int(input())
    SC = []
    SCrev = []
    for _ in range(N):
        s, c = input().split()
        c = int(c)
        SC.append((s, c))
        SCrev.append((s[::-1], c))
    dist = defaultdict(lambda: 10**18)
    q = [(c, s, "") for s, c in SC] + [(c, "", s) for s, c in SCrev]
    for c, l, r in q:
        dist[(l, r)] = min(dist[(l, r)], c)
    heapify(q)
    while q:
        dist_v, vl, vr = heappop(q)
        if (vl == vr == "") or (len(vl) and vl == vl[::-1]) or (len(vr) and vr == vr[::-1]):
            print(dist_v)
            exit()
        if dist[(vl, vr)] != dist_v:
            continue
        if len(vl) < len(vr):
            for s, c in SC:
                if len(s) <= len(vr):
                    if vr.startswith(s):
                        vr_ = vr[len(s):]
                        dist_u = dist_v + c
                        if dist_u < dist[("", vr_)]:
                            dist[("", vr_)] = dist_u
                            heappush(q, (dist_v+c, "", vr_))
                else:
                    if s.startswith(vr):
                        vl_ = s[len(vr):]
                        dist_u = dist_v + c
                        if dist_u < dist[(vl_, "")]:
                            dist[(vl_, "")] = dist_u
                            heappush(q, (dist_v+c, vl_, ""))
        else:
            for s, c in SCrev:
                if len(s) <= len(vl):
                    if vl.startswith(s):
                        vl_ = vl[len(s):]
                        dist_u = dist_v + c
                        if dist_u < dist[(vl_, "")]:
                            dist[(vl_, "")] = dist_u
                            heappush(q, (dist_v+c, vl_, ""))
                else:
                    if s.startswith(vl):
                        vr_ = s[len(vl):]
                        dist_u = dist_v + c
                        if dist_u < dist[("", vr_)]:
                            dist[("", vr_)] = dist_u
                            heappush(q, (dist_v+c, "", vr_))
    print(-1)

main()