#include<bits/stdc++.h>
using namespace std;
using Int = long long;

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

//INSERT ABOVE HERE
signed main(){
  Int n;
  cin>>n;
  vector<Int> x(n),s(n);
  for(Int i=0;i<n;i++) cin>>x[i]>>s[i];
  vector<Int> a(n+1,0);
  for(Int i=0;i<n;i++) a[i+1]=a[i]+s[i];
  
  Int ans=0,res=-1e18;
  for(Int i=0;i<n;i++){
    chmax(res,x[i]-a[i]);
    chmax(ans,(a[i+1]-x[i])+res);
  }
  cout<<ans<<endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
using Int = long long;

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

//INSERT ABOVE HERE
signed main(){
  Int n;
  cin>>n;
  vector<Int> x(n),s(n);
  for(Int i=0;i<n;i++) cin>>x[i]>>s[i];
  vector<Int> a(n+1,0);
  for(Int i=0;i<n;i++) a[i+1]=a[i]+s[i];
  
  Int ans=0,res=-1e18;
  for(Int i=0;i<n;i++){
    chmax(res,x[i]-a[i]);
    chmax(ans,(a[i+1]-x[i])+res);
  }
  cout<<ans<<endl;
  return 0;
}

```