Score : $600$ points

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 $U_iU_{i+1}\ldots U_{i+|V|-1} = V$.

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.

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.

3
a -5
ab 10
ba -20

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.

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

5

$S =$ ab achieves the maximum beauty possible.

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

-1

Note that $S$ should be a non-empty string.