#include <bits/stdc++.h>
//#include <math.h>
using namespace std;

#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i,a,b) for (int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define BIT(x,n) bitset<n>(x)
#define PI 3.14159265358979323846

typedef long long ll;
typedef pair<int,int> P;
typedef pair<P,ll> PP;

//-----------------------------------------------------------------------------

int n;
ll x[200010],s[200010];

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin>>n;
	REP(i,n) {
		cin>>x[i]>>s[i+1];
		s[i+1]+=s[i];
	}
	ll mi=LINF,ans=0;
	REP(i,n) {
		mi=min(mi,s[i]-x[i]);
		ans=max(ans,s[i+1]-x[i]-mi);
	}
	cout<<ans<<endl;

	return 0;
}

### Prompt

Your challenge is to write 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>
//#include <math.h>
using namespace std;

#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i,a,b) for (int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define BIT(x,n) bitset<n>(x)
#define PI 3.14159265358979323846

typedef long long ll;
typedef pair<int,int> P;
typedef pair<P,ll> PP;

//-----------------------------------------------------------------------------

int n;
ll x[200010],s[200010];

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin>>n;
	REP(i,n) {
		cin>>x[i]>>s[i+1];
		s[i+1]+=s[i];
	}
	ll mi=LINF,ans=0;
	REP(i,n) {
		mi=min(mi,s[i]-x[i]);
		ans=max(ans,s[i+1]-x[i]-mi);
	}
	cout<<ans<<endl;

	return 0;
}

```