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 PyPy3 (7.3.0) (judge verdict: Runtime Error)

N=int(input())
X=[list(input().split()) for i in range(N)]
for i in range(N):
  X[i][1]=int(X[i][1])
INF=10**18
def f(S):
  if len(S)<2:
    return 0
  r=INF
  T=0
  for i in range(N):
    T=S+X[i][0]
    if T==T[::-1]:
      r=min(r,X[i][1])
  for i in range(N):
    if len(S)>=len(X[i][0]):
      if S[:len(X[i][0])]==X[i][0][::-1]:
        r=min(r,f(S[len(X[i][0]):])+X[i][1])
    elif len(S)*2>len(X[i][0]):
      if X[i][0][:len(S)]==S[::-1]:
        r=min(r,f(X[i][0][::-1][:len(S)])+X[i][1])
  return r

P=INF
for i in range(N):
  P=min(P,f(X[i][0])+X[i][1])
if P!=INF:
  print(P)
else:
  print(-1)