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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
static long long n,xi[500000],si[500000],all[8000000],t,maxn;
int main(){
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>xi[i]>>si[i];
    for(int i=0;i<n;i++)
        for(int o=i;o<n;o++){
            for(int a=i;a<=o;a++)
                all[t]+=si[a];
            all[t]-=xi[o]-xi[i];
            maxn=max(maxn,all[t]);
            t++;
        }
    cout<<maxn<<endl;
    return 0;
}