Problem:
Score : 100 points

Problem Statement

In a long narrow forest stretching east-west, there are N beasts. Below, we will call the point that is p meters from the west end Point p. The i-th beast from the west (1 ≤ i ≤ N) is at Point x_i, and can be sold for s_i yen (the currency of Japan) if captured.

You will choose two integers L and R (L ≤ R), and throw a net to cover the range from Point L to Point R including both ends, [L, R]. Then, all the beasts in the range will be captured. However, the net costs R - L yen and your profit will be (the sum of s_i over all captured beasts i) - (R - L) yen.

What is the maximum profit that can be earned by throwing a net only once?

Constraints

1 ≤ N ≤ 2 × 10^5

1 ≤ x_1 < x_2 < ... < x_N ≤ 10^{15}

1 ≤ s_i ≤ 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 s_1
x_2 s_2
:
x_N s_N

Output

When the maximum profit is X yen, print the value of X.

Sample Input 1

5
10 20
40 50
60 30
70 40
90 10

Sample Output 1

90

If you throw a net covering the range [L = 40, R = 70], the second, third and fourth beasts from the west will be captured, generating the profit of s_2 + s_3 + s_4 - (R - L) = 90 yen. It is not possible to earn 91 yen or more.

Sample Input 2

5
10 2
40 5
60 3
70 4
90 1

Sample Output 2

5

The positions of the beasts are the same as Sample 1, but the selling prices dropped significantly, so you should not aim for two or more beasts. By throwing a net covering the range [L = 40, R = 40], you can earn s_2 - (R - L) = 5 yen, and this is the maximum possible profit.

Sample Input 3

4
1 100
3 200
999999999999999 150
1000000000000000 150

Sample Output 3

299

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

    private static Scanner scan;

    private static void start() {
        scan = new Scanner(java.lang.System.in);
    }

    private static void finish() {
    }

    static class T {
        long x;
        long s;

        public T(long x, long s) {
            this.x = x;
            this.s = s;
        }
    }

    public static void main(String[] args) {
        start();

        int n = scan.nextInt();
        List<T> li = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            li.add(new T(scan.nextLong(), scan.nextLong()));
        }
        li.sort((o1, o2) -> {
            if (o1.x == o2.x) return 0;
            return (o1.x < o2.x) ? -1 : 1;
        });

        long maxProfit = 0;
        long currProfit = 0;
        long l = 0;
        for (int i = 0; i < n; i++) {
            if (currProfit + li.get(i).s - (li.get(i).x - l) < li.get(i).s) {
                l = li.get(i).x;
                currProfit = li.get(i).s;
            } else {
                currProfit = currProfit + li.get(i).s - (li.get(i).x - l);
                l = li.get(i).x;
            }
//            System.out.println(currProfit + " *");
            maxProfit = Math.max(maxProfit, currProfit);
        }
        System.out.println(maxProfit);
        finish();
    }
}