// #pragma GCC target("avx")  // CPU 処理並列化
// #pragma GCC optimize("O3")  // CPU 処理並列化
// #pragma GCC optimize("unroll-loops")  // 条件処理の呼び出しを減らす
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
#include<numeric>
#include<unordered_set>
#include<complex>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const double EPS=1e-10;
const double INF=1e+10;
const double PI=acos(-1.0);
const int C_SIZE = 3100000;
const int UF_SIZE = 3100000;
namespace{
	long long fact[C_SIZE];
	long long finv[C_SIZE];
	long long inv[C_SIZE];
	long long Comb(int a,int b){
	 	if(a<b||b<0)return 0;
	 	return fact[a]*finv[b]%mod*finv[a-b]%mod;
	}
	void init_C(int n){
		fact[0]=finv[0]=inv[1]=1;
		for(int i=2;i<n;i++){
			inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
		}
		for(int i=1;i<n;i++){
			fact[i]=fact[i-1]*i%mod;
			finv[i]=finv[i-1]*inv[i]%mod;
		}
	}
	long long pw(long long a,long long b){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%mod;
			a=a*a%mod;
			b/=2;
		}
		return ret;
	}
	long long pw_mod(long long a,long long b,long long M){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%M;
			a=a*a%M;
			b/=2;
		}
		return ret;
	}
	int pw_mod_int(int a,int b,int M){
		if(a<0)return 0;
		if(b<0)return 0;
		int ret=1;
		while(b){
			if(b%2)ret=(long long)ret*a%M;
			a=(long long)a*a%M;
			b/=2;
		}
		return ret;
	}
	int ABS(int a){return max(a,-a);}
	long long ABS(long long a){return max(a,-a);}
	double ABS(double a){return max(a,-a);}
	int sig(double r) { return (r < -EPS) ? -1 : (r > +EPS) ? +1 : 0; }
	int UF[UF_SIZE];
	void init_UF(int n){
		for(int i=0;i<n;i++)UF[i]=-1;
	}
	int FIND(int a){
		if(UF[a]<0)return a;
		return UF[a]=FIND(UF[a]);
	}
	void UNION(int a,int b){
		a=FIND(a);b=FIND(b);if(a==b)return;
		if(UF[a]>UF[b])swap(a,b);
		UF[a]+=UF[b];UF[b]=a;
	}
}
// ここから編集しろ
long long x[210000];
long long y[210000];

int main(){
	int a;scanf("%d",&a);
	long long ret=0;
	for(int i=0;i<a;i++){
		scanf("%lld%lld",x+i,y+i);
		ret=max(ret,y[i]);
	}
	long long L=inf;
	long long toc=0;
	for(int i=0;i<a;i++){
		L=min(L,toc);
		toc+=y[i];

		ret=max(ret,toc-L);
		if(i<a-1)toc-=x[i+1]-x[i];
	}
	printf("%lld\n",ret);
}
### Prompt

Please create a solution in CPP 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
// #pragma GCC target("avx")  // CPU 処理並列化
// #pragma GCC optimize("O3")  // CPU 処理並列化
// #pragma GCC optimize("unroll-loops")  // 条件処理の呼び出しを減らす
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
#include<numeric>
#include<unordered_set>
#include<complex>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const double EPS=1e-10;
const double INF=1e+10;
const double PI=acos(-1.0);
const int C_SIZE = 3100000;
const int UF_SIZE = 3100000;
namespace{
	long long fact[C_SIZE];
	long long finv[C_SIZE];
	long long inv[C_SIZE];
	long long Comb(int a,int b){
	 	if(a<b||b<0)return 0;
	 	return fact[a]*finv[b]%mod*finv[a-b]%mod;
	}
	void init_C(int n){
		fact[0]=finv[0]=inv[1]=1;
		for(int i=2;i<n;i++){
			inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
		}
		for(int i=1;i<n;i++){
			fact[i]=fact[i-1]*i%mod;
			finv[i]=finv[i-1]*inv[i]%mod;
		}
	}
	long long pw(long long a,long long b){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%mod;
			a=a*a%mod;
			b/=2;
		}
		return ret;
	}
	long long pw_mod(long long a,long long b,long long M){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%M;
			a=a*a%M;
			b/=2;
		}
		return ret;
	}
	int pw_mod_int(int a,int b,int M){
		if(a<0)return 0;
		if(b<0)return 0;
		int ret=1;
		while(b){
			if(b%2)ret=(long long)ret*a%M;
			a=(long long)a*a%M;
			b/=2;
		}
		return ret;
	}
	int ABS(int a){return max(a,-a);}
	long long ABS(long long a){return max(a,-a);}
	double ABS(double a){return max(a,-a);}
	int sig(double r) { return (r < -EPS) ? -1 : (r > +EPS) ? +1 : 0; }
	int UF[UF_SIZE];
	void init_UF(int n){
		for(int i=0;i<n;i++)UF[i]=-1;
	}
	int FIND(int a){
		if(UF[a]<0)return a;
		return UF[a]=FIND(UF[a]);
	}
	void UNION(int a,int b){
		a=FIND(a);b=FIND(b);if(a==b)return;
		if(UF[a]>UF[b])swap(a,b);
		UF[a]+=UF[b];UF[b]=a;
	}
}
// ここから編集しろ
long long x[210000];
long long y[210000];

int main(){
	int a;scanf("%d",&a);
	long long ret=0;
	for(int i=0;i<a;i++){
		scanf("%lld%lld",x+i,y+i);
		ret=max(ret,y[i]);
	}
	long long L=inf;
	long long toc=0;
	for(int i=0;i<a;i++){
		L=min(L,toc);
		toc+=y[i];

		ret=max(ret,toc-L);
		if(i<a-1)toc-=x[i+1]-x[i];
	}
	printf("%lld\n",ret);
}
```