#include <bits/stdc++.h>
using namespace std;
#define int long long
using ll=long long;
using vi=vector<int>;
using pii=pair<int,int>;
#define ALL(c) begin(c),end(c)
#define RALL(c) rbegin(c),rend(c)
#define ITR(i,b,e) for(auto i=(b);i!=(e);++i)
#define FORE(x,c) for(auto &x:c)
#define REPF(i,a,n) for(int i=a,i##len=(int)(n);i<i##len;++i)
#define REP(i,n) REPF(i,0,n)
#define REPR(i,n) for(int i=(int)(n);i>=0;--i)
#define SZ(c) ((int)c.size())
#define CONTAIN(c,x) (c.find(x)!=end(c))
#define OUTOFRANGE(y,x,h,w) (y<0||x<0||y>=h||x>=w)
#define dump(...)
const int DX[9]={0,1,0,-1,1,1,-1,-1,0},DY[9]={-1,0,1,0,-1,1,1,-1,0};
#define INF (1001001001)
#define INFLL (1001001001001001001ll)
template<class T> ostream& operator << (ostream &os,const vector<T> &v) {
    ITR(i,begin(v),end(v)) os<<*i<<(i==end(v)-1?"":"\n"); return os; }
template<class T> istream& operator >> (istream &is,vector<T> &v) {
    ITR(i,begin(v),end(v)) is>>*i; return is; }
template<class T> istream& operator >> (istream &is, pair<T,T> &p) {
    is>>p.first>>p.second; return is; }
template <class T> bool chmax(T &a,const T &b) {if(a<b) {a=b;return 1;} return 0;}
template <class T> bool chmin(T &a,const T &b) {if(b<a) {a=b;return 1;} return 0;}
template <class T> using heap = priority_queue<T,vector<T>,greater<T>>;
struct before_main_function {
    before_main_function() {
        #ifdef int
            #undef INF
            #define INF INFLL
            #define stoi stoll
        #endif
        cin.tie(0);ios::sync_with_stdio(false);
        cout<<setprecision(15)<<fixed;
        #define endl "\n"
    }
} before_main_function;
//------------------8<------------------------------------8<--------------------

signed main() {
    int N;
    cin>>N;
    vector<int> s(N);
    vector<int> x(N);
    vector<int> a(N,0),b(N,0);
    REP(i,N) {
        cin>>x[i]>>s[i];
        if(i>0) s[i]+=s[i-1];
        a[i]=s[i]-x[i];
        if(i==0) b[i]=-x[i];
        else b[i]=s[i-1]-x[i];
    }
    int ans=0;
    int mn=INF;
    REP(i,N) {
        chmin(mn,b[i]);
        int tmp=a[i]-mn;
        chmax(ans,tmp);
    }
    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 <bits/stdc++.h>
using namespace std;
#define int long long
using ll=long long;
using vi=vector<int>;
using pii=pair<int,int>;
#define ALL(c) begin(c),end(c)
#define RALL(c) rbegin(c),rend(c)
#define ITR(i,b,e) for(auto i=(b);i!=(e);++i)
#define FORE(x,c) for(auto &x:c)
#define REPF(i,a,n) for(int i=a,i##len=(int)(n);i<i##len;++i)
#define REP(i,n) REPF(i,0,n)
#define REPR(i,n) for(int i=(int)(n);i>=0;--i)
#define SZ(c) ((int)c.size())
#define CONTAIN(c,x) (c.find(x)!=end(c))
#define OUTOFRANGE(y,x,h,w) (y<0||x<0||y>=h||x>=w)
#define dump(...)
const int DX[9]={0,1,0,-1,1,1,-1,-1,0},DY[9]={-1,0,1,0,-1,1,1,-1,0};
#define INF (1001001001)
#define INFLL (1001001001001001001ll)
template<class T> ostream& operator << (ostream &os,const vector<T> &v) {
    ITR(i,begin(v),end(v)) os<<*i<<(i==end(v)-1?"":"\n"); return os; }
template<class T> istream& operator >> (istream &is,vector<T> &v) {
    ITR(i,begin(v),end(v)) is>>*i; return is; }
template<class T> istream& operator >> (istream &is, pair<T,T> &p) {
    is>>p.first>>p.second; return is; }
template <class T> bool chmax(T &a,const T &b) {if(a<b) {a=b;return 1;} return 0;}
template <class T> bool chmin(T &a,const T &b) {if(b<a) {a=b;return 1;} return 0;}
template <class T> using heap = priority_queue<T,vector<T>,greater<T>>;
struct before_main_function {
    before_main_function() {
        #ifdef int
            #undef INF
            #define INF INFLL
            #define stoi stoll
        #endif
        cin.tie(0);ios::sync_with_stdio(false);
        cout<<setprecision(15)<<fixed;
        #define endl "\n"
    }
} before_main_function;
//------------------8<------------------------------------8<--------------------

signed main() {
    int N;
    cin>>N;
    vector<int> s(N);
    vector<int> x(N);
    vector<int> a(N,0),b(N,0);
    REP(i,N) {
        cin>>x[i]>>s[i];
        if(i>0) s[i]+=s[i-1];
        a[i]=s[i]-x[i];
        if(i==0) b[i]=-x[i];
        else b[i]=s[i-1]-x[i];
    }
    int ans=0;
    int mn=INF;
    REP(i,N) {
        chmin(mn,b[i]);
        int tmp=a[i]-mn;
        chmax(ans,tmp);
    }
    cout<<ans<<endl;
    return 0;
}

```