Art Exhibition

Seungwon owns $N$ artworks, numbered from 1 to $N$. Artwork $i$ has size $A_i$ and value $B_i$.

Today Seungwon wants to exhibit some of the artworks on the first floor of his mansion. He chooses the artworks to exhibit under the following condition.

• Let $A_{max}$ be the largest size and $A_{min}$ the smallest size among the exhibited artworks, and let $S$ be the sum of their values.
• The quantity $S - (A_{max} - A_{min})$ must be as large as possible.

At least one artwork is exhibited. Given the sizes and values of the $N$ artworks, write a program that finds the maximum possible value of $S - (A_{max} - A_{min})$.

The first line contains the number of artworks $N$ ($2 \le N \le 500{,}000$).

Each of the next $N$ lines contains the size $A_i$ and the value $B_i$ of one artwork, in order from artwork 1 to artwork $N$. ($1 \le A_i \le 1{,}000{,}000{,}000{,}000{,}000 = 10^{15}$, $1 \le B_i \le 1{,}000{,}000{,}000$)

Print the maximum value of $S - (A_{max} - A_{min})$ on the first line.

In the first sample, Seungwon owns 3 artworks with the following sizes and values.

• Artwork 1 has size 2 and value 3.
• Artwork 2 has size 11 and value 2.
• Artwork 3 has size 4 and value 5.

Exhibiting artworks 1 and 3 gives $S - (A_{max} - A_{min}) = 6$, which is the largest possible value.

• The largest exhibited artwork is artwork 3, so $A_{max} = 4$.
• The smallest exhibited artwork is artwork 1, so $A_{min} = 2$.
• The sum of the exhibited values is $3 + 5 = 8$.

Therefore $S = 8$ and $S - (A_{max} - A_{min})$ is 6.