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

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcount

#define INF INT_MAX/3

ll n;
ll x[200010],s[200010];
ll ss[200010];

int main(){
	cin>>n;
  rep(i,n)cin>>x[i]>>s[i];
  set<P> st;
  rep(i,n){
    ss[i+1]=ss[i]+s[i];
    st.insert(P(-(ss[i+1]-x[i]),i));
  }
  ll nx=x[0];
  ll ns=0;
  ll res=0;
  rep(i,n){
    P t=*st.begin();
    maxch(res,-t.fi+nx-ns);
    if(i+1<n)nx=x[i+1];
    ns+=s[i];
    st.erase(P(-(ss[i+1]-x[i]),i));
  }
  cout<<res<<endl;

	return 0;
}

### Prompt

Please create a solution in cpp 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;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcount

#define INF INT_MAX/3

ll n;
ll x[200010],s[200010];
ll ss[200010];

int main(){
	cin>>n;
  rep(i,n)cin>>x[i]>>s[i];
  set<P> st;
  rep(i,n){
    ss[i+1]=ss[i]+s[i];
    st.insert(P(-(ss[i+1]-x[i]),i));
  }
  ll nx=x[0];
  ll ns=0;
  ll res=0;
  rep(i,n){
    P t=*st.begin();
    maxch(res,-t.fi+nx-ns);
    if(i+1<n)nx=x[i+1];
    ns+=s[i];
    st.erase(P(-(ss[i+1]-x[i]),i));
  }
  cout<<res<<endl;

	return 0;
}

```