#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){return o<<"("<<p.fs<<","<<p.sc<<")";}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){o<<"sz = "<<vc.size()<<endl<<"[";for(const T& v:vc) o<<v<<",";o<<"]";return o;}
using ll = long long;
struct segtree{
	using D = long long;
	D inf = 1e18;

	int N;
	vector<D> val;

	segtree(){}
	segtree(int n){
		N=1;
		while(N<n) N*=2;
		val.assign(N*2,-inf);
	}
	segtree(const vector<D>& ds){
		int n = ds.size();
		N=1;
		while(N<n) N*=2;
		val.assign(N*2,-inf);
		rep(i,n) val[i+N] = ds[i];
		for(int i=N-1;i>0;i--) val[i] = max(val[i*2],val[i*2+1]);
	}
	void assign(int k,D d){
		k+=N;
		val[k]=d;
		k/=2;
		while(k){
			val[k] = max(val[k*2],val[k*2+1]);
			k/=2;
		}
	}
	D getmax(int a,int b){
		D res = -inf;
		a+=N,b+=N;
		while(a<b){
			if(a&1) chmax(res,val[a++]);
			if(b&1) chmax(res,val[--b]);
			a/=2,b/=2;
		}
		return res;
	}
};
int main(){
	int N;
	cin>>N;
	segtree seg(N+1);
	ll ans = 0;
	ll sump = 0;
	rep(i,N){
		ll x,p;
		cin>>x>>p;
		seg.assign(i,x-sump);
		sump += p;
		ll tmp = sump + seg.getmax(0,i+1) - x;
		chmax(ans,tmp);
	}
	cout<<ans<<endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){return o<<"("<<p.fs<<","<<p.sc<<")";}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){o<<"sz = "<<vc.size()<<endl<<"[";for(const T& v:vc) o<<v<<",";o<<"]";return o;}
using ll = long long;
struct segtree{
	using D = long long;
	D inf = 1e18;

	int N;
	vector<D> val;

	segtree(){}
	segtree(int n){
		N=1;
		while(N<n) N*=2;
		val.assign(N*2,-inf);
	}
	segtree(const vector<D>& ds){
		int n = ds.size();
		N=1;
		while(N<n) N*=2;
		val.assign(N*2,-inf);
		rep(i,n) val[i+N] = ds[i];
		for(int i=N-1;i>0;i--) val[i] = max(val[i*2],val[i*2+1]);
	}
	void assign(int k,D d){
		k+=N;
		val[k]=d;
		k/=2;
		while(k){
			val[k] = max(val[k*2],val[k*2+1]);
			k/=2;
		}
	}
	D getmax(int a,int b){
		D res = -inf;
		a+=N,b+=N;
		while(a<b){
			if(a&1) chmax(res,val[a++]);
			if(b&1) chmax(res,val[--b]);
			a/=2,b/=2;
		}
		return res;
	}
};
int main(){
	int N;
	cin>>N;
	segtree seg(N+1);
	ll ans = 0;
	ll sump = 0;
	rep(i,N){
		ll x,p;
		cin>>x>>p;
		seg.assign(i,x-sump);
		sump += p;
		ll tmp = sump + seg.getmax(0,i+1) - x;
		chmax(ans,tmp);
	}
	cout<<ans<<endl;
}

```