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

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

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

int sum1[222222],sum2[222222];
int N;
int X[222222],S[222222];
int sum[222222];

signed main(){
    cin>>N;
    rep(i,N){
        cin>>X[i]>>S[i];
    }

    rep(i,N)sum[i+1]=sum[i]+S[i];

    for(int i=1;i<=N;i++){
        sum1[i]=sum[i]-X[i-1];
        sum2[i]=sum[i-1]-X[i-1];
    }

    int mi=LLONG_MAX;

    int ans=0;
    for(int i=1;i<=N;i++){
        chmin(mi,sum2[i]);
        chmax(ans,sum1[i]-mi);
    }
    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;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

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

int sum1[222222],sum2[222222];
int N;
int X[222222],S[222222];
int sum[222222];

signed main(){
    cin>>N;
    rep(i,N){
        cin>>X[i]>>S[i];
    }

    rep(i,N)sum[i+1]=sum[i]+S[i];

    for(int i=1;i<=N;i++){
        sum1[i]=sum[i]-X[i-1];
        sum2[i]=sum[i-1]-X[i-1];
    }

    int mi=LLONG_MAX;

    int ans=0;
    for(int i=1;i<=N;i++){
        chmin(mi,sum2[i]);
        chmax(ans,sum1[i]-mi);
    }
    cout<<ans<<endl;
    return 0;
}

```