#include<iostream>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<iomanip>
#include<math.h>
#include<bitset>
#include<cassert>
#include<random>
#include<time.h>
using namespace std;
using ll=long long;
using ld=long double;
using P=pair<ll,ll>;
#define MOD 1000000007LL
#define INF 1000000000LL
#define EPS 1e-10
#define FOR(i,n,m) for(ll i=n;i<(ll)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v)  sort(ALL(v));v.erase(unique(ALL(v)),v.end());
#define pb push_back

/* --------------------------------------- */

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	ll n;
	cin >> n;
	vector<ll> x(n);
	vector<ll> s(n);
	REP(i,n) cin >> x[i] >> s[i];
	ll ans = 0;
	REP(i,n) ans = max(ans, s[i]);
	priority_queue<ll> q;
	q.push(0);
	ll sum = 0;
	REP(i,n) {
		q.push(x[i] - x[0] - sum);
		sum += s[i];
		ans = max(ans, sum - (x[i] - x[0]) + q.top());
	}
	cout << ans << endl;
	return 0;
}

/* --------------------------------------- */

### Prompt

Create a solution in CPP for 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<iostream>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<iomanip>
#include<math.h>
#include<bitset>
#include<cassert>
#include<random>
#include<time.h>
using namespace std;
using ll=long long;
using ld=long double;
using P=pair<ll,ll>;
#define MOD 1000000007LL
#define INF 1000000000LL
#define EPS 1e-10
#define FOR(i,n,m) for(ll i=n;i<(ll)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v)  sort(ALL(v));v.erase(unique(ALL(v)),v.end());
#define pb push_back

/* --------------------------------------- */

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	ll n;
	cin >> n;
	vector<ll> x(n);
	vector<ll> s(n);
	REP(i,n) cin >> x[i] >> s[i];
	ll ans = 0;
	REP(i,n) ans = max(ans, s[i]);
	priority_queue<ll> q;
	q.push(0);
	ll sum = 0;
	REP(i,n) {
		q.push(x[i] - x[0] - sum);
		sum += s[i];
		ans = max(ans, sum - (x[i] - x[0]) + q.top());
	}
	cout << ans << endl;
	return 0;
}

/* --------------------------------------- */

```