#include <bits/stdc++.h>
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)

using namespace std;
const ll MAX_N=2e5;
ll N,X[MAX_N],S[MAX_N],psm[MAX_N+1];
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>N;
	REP(i,N)cin>>X[i]>>S[i];
	REP(i,N)psm[i+1]=psm[i]+S[i];
	priority_queue<ll> pque;
	ll ans=0;
	REP(i,N){
		pque.push(X[i]-psm[i]);
		ans=max(ans,psm[i+1]-X[i]+pque.top());
	}
	cout<<ans<<endl;
}
### Prompt

Generate 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>
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)

using namespace std;
const ll MAX_N=2e5;
ll N,X[MAX_N],S[MAX_N],psm[MAX_N+1];
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>N;
	REP(i,N)cin>>X[i]>>S[i];
	REP(i,N)psm[i+1]=psm[i]+S[i];
	priority_queue<ll> pque;
	ll ans=0;
	REP(i,N){
		pque.push(X[i]-psm[i]);
		ans=max(ans,psm[i+1]-X[i]+pque.top());
	}
	cout<<ans<<endl;
}
```