#include <iostream>
#include <algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<iomanip>
#define sqr(x) (x)*(x)
using namespace std;
long long n,m,i,j,dp[200005],x[200005],val[200005],all,ans=-1e17;
int main()
{
	ios::sync_with_stdio(0);
	cin>>n;
	for (i=1;i<=n;i++)
	{
		cin>>x[i]>>val[i];
	}
	for (i=1;i<=n;i++)
	{
		dp[i]=max(dp[i-1]-(x[i]-x[i-1]),0ll)+val[i];
		ans=max(ans,dp[i]);
	}
	cout<<ans;
	return 0;
}
### Prompt

In cpp, your task is to solve 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 <algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<iomanip>
#define sqr(x) (x)*(x)
using namespace std;
long long n,m,i,j,dp[200005],x[200005],val[200005],all,ans=-1e17;
int main()
{
	ios::sync_with_stdio(0);
	cin>>n;
	for (i=1;i<=n;i++)
	{
		cin>>x[i]>>val[i];
	}
	for (i=1;i<=n;i++)
	{
		dp[i]=max(dp[i-1]-(x[i]-x[i-1]),0ll)+val[i];
		ans=max(ans,dp[i]);
	}
	cout<<ans;
	return 0;
}
```