#include <bits/stdc++.h>
#define int long long
using namespace std;

typedef pair<int,int> P;

signed main(){
	int n,ma = 0,dp[200010] = {};
	vector<P> vec;
	cin >> n;
	for(int i = 0;i < n;i++){
		int x,s;
		cin >> x >> s;
		vec.push_back(P(x,s));
	}
	sort(vec.begin(),vec.end());
	for(int i = 0;i < n;i++){
		if(!i) dp[i] = vec[i].second;
		else dp[i] = max(vec[i].second,dp[i - 1] + vec[i].second - vec[i].first + vec[i - 1].first);
		ma = max(ma,dp[i]);
	}
	cout << ma << endl;
	return 0;
}
### 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>
#define int long long
using namespace std;

typedef pair<int,int> P;

signed main(){
	int n,ma = 0,dp[200010] = {};
	vector<P> vec;
	cin >> n;
	for(int i = 0;i < n;i++){
		int x,s;
		cin >> x >> s;
		vec.push_back(P(x,s));
	}
	sort(vec.begin(),vec.end());
	for(int i = 0;i < n;i++){
		if(!i) dp[i] = vec[i].second;
		else dp[i] = max(vec[i].second,dp[i - 1] + vec[i].second - vec[i].first + vec[i - 1].first);
		ma = max(ma,dp[i]);
	}
	cout << ma << endl;
	return 0;
}
```