#include <algorithm>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <sstream>
#include <functional>
#include <map>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <list>
#include <numeric>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> P;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const ll INF = 1LL<<29;
const ll mod = 1e9+7;
#define rep(i,n) for(int (i)=0;(i)<(ll)(n);++(i))
#define repd(i,n,d) for(ll (i)=0;(i)<(ll)(n);(i)+=(d))
#define all(v) (v).begin(), (v).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset((m),(v),sizeof(m))
#define chmin(X,Y) ((X)>(Y)?X=(Y),true:false)
#define chmax(X,Y) ((X)<(Y)?X=(Y),true:false)
#define fst first
#define snd second
#define UNIQUE(x) (x).erase(unique(all(x)),(x).end())
template<class T> ostream &operator<<(ostream &os, const vector<T> &v){int n=v.size();rep(i,n)os<<v[i]<<(i==n-1?"":" ");return os;}

#define N 200010

ll x[N], s[N], s2[N], f[N];

int main(){
	int n;
	cin>>n;
	for(int i = 1; i <= n; i++){
		cin>>x[i]>>s[i];
	}
	x[n+1] = 1e15+1;
	for(int i = 1; i <= n; i++) s2[i] = s[i]+x[i]-x[i+1];
	for(int i = 1; i <= n; i++) f[i] = max(0LL, f[i-1])+s2[i];
	ll res = 0;
	for(int i = 1; i <= n; i++) res = max(res, f[i]-x[i]+x[i+1]);
	cout<<res<<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 <algorithm>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <sstream>
#include <functional>
#include <map>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <list>
#include <numeric>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> P;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const ll INF = 1LL<<29;
const ll mod = 1e9+7;
#define rep(i,n) for(int (i)=0;(i)<(ll)(n);++(i))
#define repd(i,n,d) for(ll (i)=0;(i)<(ll)(n);(i)+=(d))
#define all(v) (v).begin(), (v).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset((m),(v),sizeof(m))
#define chmin(X,Y) ((X)>(Y)?X=(Y),true:false)
#define chmax(X,Y) ((X)<(Y)?X=(Y),true:false)
#define fst first
#define snd second
#define UNIQUE(x) (x).erase(unique(all(x)),(x).end())
template<class T> ostream &operator<<(ostream &os, const vector<T> &v){int n=v.size();rep(i,n)os<<v[i]<<(i==n-1?"":" ");return os;}

#define N 200010

ll x[N], s[N], s2[N], f[N];

int main(){
	int n;
	cin>>n;
	for(int i = 1; i <= n; i++){
		cin>>x[i]>>s[i];
	}
	x[n+1] = 1e15+1;
	for(int i = 1; i <= n; i++) s2[i] = s[i]+x[i]-x[i+1];
	for(int i = 1; i <= n; i++) f[i] = max(0LL, f[i-1])+s2[i];
	ll res = 0;
	for(int i = 1; i <= n; i++) res = max(res, f[i]-x[i]+x[i+1]);
	cout<<res<<endl;
	return 0;
}

```