Problem9805--ABC264 —— G - String Fair

9805: ABC264 —— G - String Fair

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Time Limit : 1.000 sec  Memory Limit : 512 MiB

Description

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 1 Input

3
a -5
ab 10
ba -20

Sample 1 Output

Infinity

For example, if S= abzabz:

  • The score determined by the 1-st criterion is 2×(−5)=−10 points, because a occurs twice in S as a consecutive subsequence.
  • The score determined by the 2-nd criterion is 2×10=20 points, because ab occurs twice in S as a consecutive subsequence.
  • The score determined by the 3-rd criterion is 0×(−20)=00 points, because ba occurs 00 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×(−5)=−15 points, because a occurs 3 times in S as a consecutive subsequence.
  • The score determined by the 2-nd criterion is 3×10=30 points, because ab occurs 3 times in S as a consecutive subsequence.
  • The score determined by the 3-rd criterion is 0×(−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 2 Input

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 2 Output

5
S= ab achieves the maximum beauty possible.

Sample 3 Input

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 3 Output

-1

Note that S should be a non-empty string.

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