#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<vector>
#include<queue>
#include<map>
#include<tuple>
#include<bitset>
#include<iomanip>
#define rep(index,num) for(int index=0;index<num;index++)
#define rep1(index,num) for(int index=1;index<=num;index++)
#define scan(argument) cin>>argument
#define prin(argument) cout<<argument<<endl
#define kaigyo cout<<endl
#define eps 1e-15
#define mp(a1,a2) make_pair(a1,a2)
typedef long long ll;
using namespace std;
typedef pair<ll,ll> pll;
typedef pair<int,int> pint;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<pint> vpint;
typedef vector<pll> vpll;
ll INFl=1e+18+1;
int INF=1e+9+1;
int main(){
	int N;
	ll x[200001],s[200001];
	ll ruisekiwa[400001]={};
	ll ruisekimin=0;
	ll ans=0;
	scan(N);
	rep(i,N){
		scan(x[i]);scan(s[i]);
	}
	rep1(i,2*N-1){
		if(i%2){
			ruisekiwa[i]=ruisekiwa[i-1]+s[i/2];
		}
		else{
			ruisekiwa[i]=ruisekiwa[i-1]-(x[i/2]-x[i/2-1]);
		}
		ruisekimin=min(ruisekimin,ruisekiwa[i]);
		ans=max(ans,ruisekiwa[i]-ruisekimin);
	}
	prin(ans);
	return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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.

Examples

Input

5
10 20
40 50
60 30
70 40
90 10


Output

90


Input

5
10 2
40 5
60 3
70 4
90 1


Output

5


Input

4
1 100
3 200
999999999999999 150
1000000000000000 150


Output

299

### Response

```cpp
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<vector>
#include<queue>
#include<map>
#include<tuple>
#include<bitset>
#include<iomanip>
#define rep(index,num) for(int index=0;index<num;index++)
#define rep1(index,num) for(int index=1;index<=num;index++)
#define scan(argument) cin>>argument
#define prin(argument) cout<<argument<<endl
#define kaigyo cout<<endl
#define eps 1e-15
#define mp(a1,a2) make_pair(a1,a2)
typedef long long ll;
using namespace std;
typedef pair<ll,ll> pll;
typedef pair<int,int> pint;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<pint> vpint;
typedef vector<pll> vpll;
ll INFl=1e+18+1;
int INF=1e+9+1;
int main(){
	int N;
	ll x[200001],s[200001];
	ll ruisekiwa[400001]={};
	ll ruisekimin=0;
	ll ans=0;
	scan(N);
	rep(i,N){
		scan(x[i]);scan(s[i]);
	}
	rep1(i,2*N-1){
		if(i%2){
			ruisekiwa[i]=ruisekiwa[i-1]+s[i/2];
		}
		else{
			ruisekiwa[i]=ruisekiwa[i-1]-(x[i/2]-x[i/2-1]);
		}
		ruisekimin=min(ruisekimin,ruisekiwa[i]);
		ans=max(ans,ruisekiwa[i]-ruisekimin);
	}
	prin(ans);
	return 0;
}

```