
#include "bits/stdc++.h"
using namespace std;

#define DEBUG(x) cout<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cout<<#v<<":";for(int i=0;i<v.size();i++) cout<<" "<<v[i]; cout<<endl

typedef long long ll;
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
const int inf = 1000000001;
const ll INF = 2e18 * 2;
#define MOD 1000000007
#define mod 1000000009
#define pi 3.14159265358979323846
#define Sp(p) cout<<setprecision(15)<< fixed<<p<<endl;
int dx[4] = { 1,0,-1,0 }, dy[4] = { 0,1,0,-1 };
int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };


int main() {
	int n, i, j;
	cin >> n;
	vl x(n), s(n);
	for (i = 0; i < n; i++) {
		cin >> x[i] >> s[i];
	}
	vl sum(n + 1);
	vl left(n), right(n);
	sum[0] = 0;
	for (i = 0; i < n; i++) {
		sum[i + 1] = sum[i] + s[i];
		left[i] = sum[i] - x[i];
		right[i] = sum[i + 1] - x[i];
	}
	/*
	DEBUG_VEC(left);
	DEBUG_VEC(right);
	*/
	ll ans = -INF;
	deque<ll> dqr;
	for (i = 0; i < n; i++) {
		while (!dqr.empty() && right[dqr.back()] < right[i]) {
			dqr.pop_back();
		}
		dqr.push_back(i);
	}
	

	for (i = 0; i < n; i++) {
		/*
		for (j = 0; j < dqr.size(); j++) {
			cout << dqr[j] << " ";
		}
		cout << endl;
		*/
		ans = max(ans, right[dqr[0]] - left[i]);
		while (!dqr.empty() && dqr.front() <= i) {
			dqr.pop_front();
		}
	}
	cout << ans << endl;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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"
using namespace std;

#define DEBUG(x) cout<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cout<<#v<<":";for(int i=0;i<v.size();i++) cout<<" "<<v[i]; cout<<endl

typedef long long ll;
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
const int inf = 1000000001;
const ll INF = 2e18 * 2;
#define MOD 1000000007
#define mod 1000000009
#define pi 3.14159265358979323846
#define Sp(p) cout<<setprecision(15)<< fixed<<p<<endl;
int dx[4] = { 1,0,-1,0 }, dy[4] = { 0,1,0,-1 };
int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };


int main() {
	int n, i, j;
	cin >> n;
	vl x(n), s(n);
	for (i = 0; i < n; i++) {
		cin >> x[i] >> s[i];
	}
	vl sum(n + 1);
	vl left(n), right(n);
	sum[0] = 0;
	for (i = 0; i < n; i++) {
		sum[i + 1] = sum[i] + s[i];
		left[i] = sum[i] - x[i];
		right[i] = sum[i + 1] - x[i];
	}
	/*
	DEBUG_VEC(left);
	DEBUG_VEC(right);
	*/
	ll ans = -INF;
	deque<ll> dqr;
	for (i = 0; i < n; i++) {
		while (!dqr.empty() && right[dqr.back()] < right[i]) {
			dqr.pop_back();
		}
		dqr.push_back(i);
	}
	

	for (i = 0; i < n; i++) {
		/*
		for (j = 0; j < dqr.size(); j++) {
			cout << dqr[j] << " ";
		}
		cout << endl;
		*/
		ans = max(ans, right[dqr[0]] - left[i]);
		while (!dqr.empty() && dqr.front() <= i) {
			dqr.pop_front();
		}
	}
	cout << ans << endl;
}
```