Problem Statement

In a string fair, they determine the beauty of a non-empty string S consisting of lowercase English letters.
The beauty of string S equals the sum of N scores determined by N criteria.
For $i = 1, 2, \ldots, N$, the score determined by the i-th criterion is
"the number of occurrences of a string T_i (of length at most 3 given in the Input) in S as a consecutive subsequence" multiplied by P_i.
Print the maximum possible beauty of a non-empty string S consisting of lowercase English letters.
If it is possible to obtain infinitely large beauty, print Infinity instead.
Here, the number of occurrences of a string V in $a string U = U_1U_2\ldots U_{|U|}$ as a consecutive subsequence is defined to be
the number of integers i such that $1 \leq i \leq |U|-|V|+1$ and $U_iU_{i+1}\ldots U_{i+|V|-1} = V$.

Input

Input is given from Standard Input in the following format:
N
T_1 P_1
T_2 P_2
$\vdots$
T_N P_N

Output

Print the maximum possible beauty of a non-empty string S consisting of lowercase English letters.
If it is possible to obtain infinitely large beauty, print Infinity instead.

Constraints

• $1 \leq N \leq 18278$
• N is an integer.
• T_i is a string of length between 1 and 3 consisting of lowercase English letters.
• $i \neq j \Rightarrow T_i \neq T_j$
• -$10^9 \leq P_i \leq 10^9$
• P_i is an integer.

Sample Input 1

3
a -5
ab 10
ba -20

Sample Output 1

Infinity

For example, if S = abzabz:

• The score determined by the 1-st criterion is $2 \times (-5) = -10 points, because a$ occurs twice in S as a consecutive subsequence.
• The score determined by the 2-nd criterion is $2 \times 10 = 20$ points, because ab occurs twice in S as a consecutive subsequence.
• The score determined by the 3-rd criterion is $0 \times (-20) = 0 points, because ba occurs 0$ times in S as a consecutive subsequence.

Thus, the beauty of S is (-10) + 20 + 0 = 10.
As another example, if S = abzabzabz:

• The score determined by the 1-st criterion is $3 \times (-5) = -15 points, because a occurs 3$ times in S as a consecutive subsequence.
• The score determined by the 2-nd criterion is $3 \times 10 = 30 points, because ab occurs 3$ times in S as a consecutive subsequence.
• The score determined by the 3-rd criterion is $0 \times (-20) = 0 points, because ba occurs 0$ times in S as a consecutive subsequence.

Thus, the beauty of S is (-15) + 30 + 0 = 15.
In general, for a positive integer X, if S is a concatenation of X copies of abz, then the beauty of S is 5X.
Since you can obtain as large beauty as you want, Infinity should be printed.

Sample Input 2

28
a -5
ab 10
ba -20
bb -20
bc -20
bd -20
be -20
bf -20
bg -20
bh -20
bi -20
bj -20
bk -20
bl -20
bm -20
bn -20
bo -20
bp -20
bq -20
br -20
bs -20
bt -20
bu -20
bv -20
bw -20
bx -20
by -20
bz -20

Sample Output 2

5

S = ab achieves the maximum beauty possible.

Sample Input 3

26
a -1
b -1
c -1
d -1
e -1
f -1
g -1
h -1
i -1
j -1
k -1
l -1
m -1
n -1
o -1
p -1
q -1
r -1
s -1
t -1
u -1
v -1
w -1
x -1
y -1
z -1

Sample Output 3

-1

Note that S should be a non-empty string.

见题面 Input 部分

见题面 Output 部分