#include<bits/stdc++.h>
using namespace std;
struct node{
    long long x,s;
}a[200005];
long long n,maxn,sum;
long long f[200005];
bool cmp(node a,node b){
    return a.s<b.s;
}
int main(){
    cin>>n;
    for(int i=1;i<=n;i++)
    cin>>a[i].s>>a[i].x;
    sort(a+1,a+n+1,cmp);
    f[1]=maxn=a[1].x;
//  for(int i=2;i<=n;i++){
        for(int j=2;j<=n;j++){
            f[j]=max(f[j-1]+a[j].x-a[j].s+a[j-1].s,a[j].x);
        }
//  }
    for(int i=1;i<=n;i++){
        maxn=max(maxn,f[i]);
    }
    cout<<maxn<<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>
using namespace std;
struct node{
    long long x,s;
}a[200005];
long long n,maxn,sum;
long long f[200005];
bool cmp(node a,node b){
    return a.s<b.s;
}
int main(){
    cin>>n;
    for(int i=1;i<=n;i++)
    cin>>a[i].s>>a[i].x;
    sort(a+1,a+n+1,cmp);
    f[1]=maxn=a[1].x;
//  for(int i=2;i<=n;i++){
        for(int j=2;j<=n;j++){
            f[j]=max(f[j-1]+a[j].x-a[j].s+a[j-1].s,a[j].x);
        }
//  }
    for(int i=1;i<=n;i++){
        maxn=max(maxn,f[i]);
    }
    cout<<maxn<<endl;
    return 0;
}
```