#include <iostream>
#include <string>
#include <vector>
#include <climits>
#include <iomanip>
#include <algorithm>
#include <tuple>
#include <queue>
#include <set>
#include <math.h>
#include <array>
#include <list>
#include <set>
#define MOD 1000000007
#define INF 1000000000
using namespace std;
typedef long long ll;
typedef tuple<ll,int,int> p;
typedef vector<int> vi;
typedef vector<ll> vl;
set <int> st;
vector<int> videos(105);
double now=0;
queue<int> num;
ll n,k,ans=-1;
ll s[200005];
ll ss[200005];
ll x[200005];
ll a[200005];
ll b[200005];
ll bmin=2e18;
int main(int argc,char const* argv[])
{
  cin >> n; 
  for(int i=0;i<n;i++)
    {
      if(i!=0) ss[i]+=ss[i-1];
      cin >> x[i] >> s[i];
      ss[i]+=s[i];
      a[i]=-x[i];
      b[i]=-x[i];
    }
  for(int i=0;i<n;i++)
    {
      if(i!=0) 
	{
	  a[i]+=ss[i-1];
	}
      b[i]=a[i];
      a[i]+=s[i];
    }
  for(int i=0;i<n;i++)
    {
      if(b[i]<bmin) bmin=b[i];
      if(a[i]-bmin>ans) ans=a[i]-bmin;
    }

  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in CPP 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 <iostream>
#include <string>
#include <vector>
#include <climits>
#include <iomanip>
#include <algorithm>
#include <tuple>
#include <queue>
#include <set>
#include <math.h>
#include <array>
#include <list>
#include <set>
#define MOD 1000000007
#define INF 1000000000
using namespace std;
typedef long long ll;
typedef tuple<ll,int,int> p;
typedef vector<int> vi;
typedef vector<ll> vl;
set <int> st;
vector<int> videos(105);
double now=0;
queue<int> num;
ll n,k,ans=-1;
ll s[200005];
ll ss[200005];
ll x[200005];
ll a[200005];
ll b[200005];
ll bmin=2e18;
int main(int argc,char const* argv[])
{
  cin >> n; 
  for(int i=0;i<n;i++)
    {
      if(i!=0) ss[i]+=ss[i-1];
      cin >> x[i] >> s[i];
      ss[i]+=s[i];
      a[i]=-x[i];
      b[i]=-x[i];
    }
  for(int i=0;i<n;i++)
    {
      if(i!=0) 
	{
	  a[i]+=ss[i-1];
	}
      b[i]=a[i];
      a[i]+=s[i];
    }
  for(int i=0;i<n;i++)
    {
      if(b[i]<bmin) bmin=b[i];
      if(a[i]-bmin>ans) ans=a[i]-bmin;
    }

  cout << ans << endl;
  return 0;
}

```