#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll>P;

P p[600000];
ll sum[600000];
ll D[600000];
int main(){
	int n;scanf("%d",&n);
	rep(i,n){
		scanf("%lld%lld",&p[i].first,&p[i].second);
	}
	sort(p,p+n);
	rep(i,n){
		if(i)sum[i]+=sum[i-1];
		sum[i]+=p[i].second;
	}
	D[n-1]=LLONG_MIN/2;
	ll ans=0;
	for(int i=n-1;i>=0;i--){
		if(i!=n-1){
			D[i]=D[i+1];
		}
		D[i]=max(D[i],sum[i]-p[i].first);
		ll cnt=D[i];
		if(i)cnt-=sum[i-1];
		ans=max(ans,cnt+p[i].first);
	}
	cout<<ans<<endl;
}

### Prompt

Generate a Cpp solution to 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 <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll>P;

P p[600000];
ll sum[600000];
ll D[600000];
int main(){
	int n;scanf("%d",&n);
	rep(i,n){
		scanf("%lld%lld",&p[i].first,&p[i].second);
	}
	sort(p,p+n);
	rep(i,n){
		if(i)sum[i]+=sum[i-1];
		sum[i]+=p[i].second;
	}
	D[n-1]=LLONG_MIN/2;
	ll ans=0;
	for(int i=n-1;i>=0;i--){
		if(i!=n-1){
			D[i]=D[i+1];
		}
		D[i]=max(D[i],sum[i]-p[i].first);
		ll cnt=D[i];
		if(i)cnt-=sum[i-1];
		ans=max(ans,cnt+p[i].first);
	}
	cout<<ans<<endl;
}

```