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: Accepted)

import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")


n = int(input())
from collections import defaultdict
ns = defaultdict(set)
sc = [None]*n
nodes = []
for i in range(n):
    s,c = input().split()
    c = int(c)
    sc[i] = (s,c)
    nodes.extend([(0,s[i:]) for i in range(len(s))])
    nodes.extend([(1,s[:i]) for i in range(1, len(s)+1)])

for u in nodes:
    for s,c in sc:
        if u[0]==0 and s.endswith(u[1][::-1]):
            v = (1, s[:-len(u[1])])
            ns[u].add((c,v))
        elif u[0]==0 and u[1][::-1].endswith(s):
            v = (0, u[1][len(s):])
            ns[u].add((c,v))
        elif u[0]==1 and s.startswith(u[1][::-1]):
            v = (0, s[len(u[1]):])
            ns[u].add((c,v))
        elif u[0]==1 and u[1][::-1].startswith(s):
            v = (1, u[1][:-len(s)])
            ns[u].add((c,v))
start = 0
for s,c in sc:
    ns[start].add((c, (0,s)))
    ns[start].add((c, (1,s)))
def dijkstra(start):
    import heapq
    vals = {}
    h = [(0, start)] # (距離, ノード番号)
    vals[start] = 0
    while h:
        val, u = heapq.heappop(h)
        for d, v in ns[u]:
            if v not in vals or vals[v]>val+d:
                vals[v] = val+d
                heapq.heappush(h, (vals[v], v))
    return vals
vals = dijkstra(start)
ans = float("inf")
for k,v in vals.items():
    if k==0:
        continue
    if k[1]==k[1][::-1]:
        ans = min(ans, v)
if ans==float("inf"):
    ans = -1
print(ans)