#include <bits/stdc++.h>
using namespace std;
#define F first
#define S second
#define R cin>>
#define Z class
#define ll long long
#define ln cout<<'\n'
#define in(a) insert(a)
#define pb(a) push_back(a)
#define pd(a) printf("%.10f\n",a)
#define mem(a) memset(a,0,sizeof(a))
#define all(c) (c).begin(),(c).end()
#define iter(c) __typeof((c).begin())
#define rrep(i,n) for(ll i=(ll)(n)-1;i>=0;i--)
#define REP(i,m,n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define rep(i,n) REP(i,0,n)
#define tr(it,c) for(iter(c) it=(c).begin();it!=(c).end();it++)
template<Z A>void pr(A a){cout<<a;ln;}
template<Z A,Z B>void pr(A a,B b){cout<<a<<' ';pr(b);}
template<Z A,Z B,Z C>void pr(A a,B b,C c){cout<<a<<' ';pr(b,c);}
template<Z A,Z B,Z C,Z D>void pr(A a,B b,C c,D d){cout<<a<<' ';pr(b,c,d);}
template<Z A>void PR(A a,ll n){rep(i,n){if(i)cout<<' ';cout<<a[i];}ln;}
ll check(ll n,ll m,ll x,ll y){return x>=0&&x<n&&y>=0&&y<m;}
const ll MAX=1000000007,MAXL=1LL<<60,dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
typedef pair<ll,ll> P;
int N=1<<18;
class StarrySkyTree{
public:
  ll Ma[555555],A[555555];
  void init(){memset(Ma,0,sizeof(Ma)),memset(A,0,sizeof(A));}
  void add(int a,int b,ll x,int k=0,int l=0,int r=N) {
    if(r<=a||b<=l) return;
    if(a<=l&&r<=b){
      A[k]+=x;
      while(k){
        k=(k-1)/2;
        Ma[k]=max(Ma[k*2+1]+A[k*2+1],Ma[k*2+2]+A[k*2+2]);
      }
      return;
    }
    add(a,b,x,k*2+1,l,(l+r)/2);
    add(a,b,x,k*2+2,(l+r)/2,r);
  }
  ll getMax(int a,int b,int k=0,int l=0,int r=N) {
    if(r<=a||b<=l)return -MAXL;if(a<=l&&r<=b)return Ma[k]+A[k];
    ll left=getMax(a,b,k*2+1,l,(l+r)/2),right=getMax(a,b,k*2+2,(l+r)/2,r);
    return max(left,right)+A[k];
  }
};

StarrySkyTree t;
void Main() {
  t.init();
  int n;
  R n;
  P a[n];
  rep(i,n) cin >> a[i].F >> a[i].S;
  ll ans=0;
  rep(i,n) {
    if(i) t.add(0,i,-(a[i].F-a[i-1].F));
    t.add(0,i+1,a[i].S);
    ans=max(ans,t.getMax(0,i+1));
  }
  pr(ans);
}

int main(){ios::sync_with_stdio(0);cin.tie(0);Main();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;
#define F first
#define S second
#define R cin>>
#define Z class
#define ll long long
#define ln cout<<'\n'
#define in(a) insert(a)
#define pb(a) push_back(a)
#define pd(a) printf("%.10f\n",a)
#define mem(a) memset(a,0,sizeof(a))
#define all(c) (c).begin(),(c).end()
#define iter(c) __typeof((c).begin())
#define rrep(i,n) for(ll i=(ll)(n)-1;i>=0;i--)
#define REP(i,m,n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define rep(i,n) REP(i,0,n)
#define tr(it,c) for(iter(c) it=(c).begin();it!=(c).end();it++)
template<Z A>void pr(A a){cout<<a;ln;}
template<Z A,Z B>void pr(A a,B b){cout<<a<<' ';pr(b);}
template<Z A,Z B,Z C>void pr(A a,B b,C c){cout<<a<<' ';pr(b,c);}
template<Z A,Z B,Z C,Z D>void pr(A a,B b,C c,D d){cout<<a<<' ';pr(b,c,d);}
template<Z A>void PR(A a,ll n){rep(i,n){if(i)cout<<' ';cout<<a[i];}ln;}
ll check(ll n,ll m,ll x,ll y){return x>=0&&x<n&&y>=0&&y<m;}
const ll MAX=1000000007,MAXL=1LL<<60,dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
typedef pair<ll,ll> P;
int N=1<<18;
class StarrySkyTree{
public:
  ll Ma[555555],A[555555];
  void init(){memset(Ma,0,sizeof(Ma)),memset(A,0,sizeof(A));}
  void add(int a,int b,ll x,int k=0,int l=0,int r=N) {
    if(r<=a||b<=l) return;
    if(a<=l&&r<=b){
      A[k]+=x;
      while(k){
        k=(k-1)/2;
        Ma[k]=max(Ma[k*2+1]+A[k*2+1],Ma[k*2+2]+A[k*2+2]);
      }
      return;
    }
    add(a,b,x,k*2+1,l,(l+r)/2);
    add(a,b,x,k*2+2,(l+r)/2,r);
  }
  ll getMax(int a,int b,int k=0,int l=0,int r=N) {
    if(r<=a||b<=l)return -MAXL;if(a<=l&&r<=b)return Ma[k]+A[k];
    ll left=getMax(a,b,k*2+1,l,(l+r)/2),right=getMax(a,b,k*2+2,(l+r)/2,r);
    return max(left,right)+A[k];
  }
};

StarrySkyTree t;
void Main() {
  t.init();
  int n;
  R n;
  P a[n];
  rep(i,n) cin >> a[i].F >> a[i].S;
  ll ans=0;
  rep(i,n) {
    if(i) t.add(0,i,-(a[i].F-a[i-1].F));
    t.add(0,i+1,a[i].S);
    ans=max(ans,t.getMax(0,i+1));
  }
  pr(ans);
}

int main(){ios::sync_with_stdio(0);cin.tie(0);Main();return 0;}

```