#include <bits/stdc++.h>

#define rep(i,n) for(int i=0; i<(n); i++)
#define reps(i,x,n) for(int i=x; i<(n); i++)
#define rrep(i,n) for(int i=(n)-1; i>=0; i--)
#define all(X) (X).begin(),(X).end()
#define X first
#define Y second
#define pb push_back
#define eb emplace_back

using namespace std;
typedef long long int ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

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 (a>b) { a=b; return 1; } return 0; }

template<class A, size_t N, class T> void Fill(A (&a)[N], const T &v){ fill( (T*)a, (T*)(a+N), v ); }

const ll INF = 1e9+7;



template <typename T>
class RMQ {
	T query(int l, int r, int a, int b, int k){
		if( l <= a && b <= r ){
			return seg[k];
		}else if( l < b && a < r ){
			int m = (b+a)/2;
			T vl = query(l,r, a, m, k*2);
			T vr = query(l,r, m, b, k*2+1);
			return min(vl, vr);	// merge
		}
		return INF;	// default value
	}
public:
	vector<T> seg;	// セグメントツリー (親:k/2  子:k*2,k*2+1)
	int N;
	RMQ(int size){
		N = 1;
		while( N < size ) N<<=1;
		seg.assign(N*2, INF);		// default value
	}
	void update(int pos, T v){
		for(int k=N+pos; k>0; k/=2){
			seg[k] = v;
			v = min(seg[k], seg[k^1]);	// merge
		}
	}
	// [l,r)
	T get(int l, int r){
		return query(l, r, 0, N, 1);
	}
	T get(int pos){		// =get(pos, pos+1)
		return seg[N+pos];
	}
};

int main(){
	//ios_base::sync_with_stdio(false);
	ll N, ans=0;
	ll x[200005], s[200005];

	cin >> N;
	rep(i,N) cin >> x[i] >> s[i];
	x[N] = x[N-1];

	rep(i,N) chmax(ans, s[i]);

//	set<ll> st;
//	map<ll,ll> mp;
//	rep(i,N) st.insert(x[i]);
//	for(auto t: st) mp[t] = mp.size();

	RMQ<ll> sg(200005);
	ll sum = 0;
	sg.update(0,0);
	rep(i,N){
		sum += s[i];
		ll mn  = sg.get(0,i+1);
		chmax(ans, sum - mn);
		ll d = x[i+1] - x[i];
		sum -= d;
		sg.update(i+1, sum);
	}

	cout << ans << endl;

	return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>

#define rep(i,n) for(int i=0; i<(n); i++)
#define reps(i,x,n) for(int i=x; i<(n); i++)
#define rrep(i,n) for(int i=(n)-1; i>=0; i--)
#define all(X) (X).begin(),(X).end()
#define X first
#define Y second
#define pb push_back
#define eb emplace_back

using namespace std;
typedef long long int ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

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 (a>b) { a=b; return 1; } return 0; }

template<class A, size_t N, class T> void Fill(A (&a)[N], const T &v){ fill( (T*)a, (T*)(a+N), v ); }

const ll INF = 1e9+7;



template <typename T>
class RMQ {
	T query(int l, int r, int a, int b, int k){
		if( l <= a && b <= r ){
			return seg[k];
		}else if( l < b && a < r ){
			int m = (b+a)/2;
			T vl = query(l,r, a, m, k*2);
			T vr = query(l,r, m, b, k*2+1);
			return min(vl, vr);	// merge
		}
		return INF;	// default value
	}
public:
	vector<T> seg;	// セグメントツリー (親:k/2  子:k*2,k*2+1)
	int N;
	RMQ(int size){
		N = 1;
		while( N < size ) N<<=1;
		seg.assign(N*2, INF);		// default value
	}
	void update(int pos, T v){
		for(int k=N+pos; k>0; k/=2){
			seg[k] = v;
			v = min(seg[k], seg[k^1]);	// merge
		}
	}
	// [l,r)
	T get(int l, int r){
		return query(l, r, 0, N, 1);
	}
	T get(int pos){		// =get(pos, pos+1)
		return seg[N+pos];
	}
};

int main(){
	//ios_base::sync_with_stdio(false);
	ll N, ans=0;
	ll x[200005], s[200005];

	cin >> N;
	rep(i,N) cin >> x[i] >> s[i];
	x[N] = x[N-1];

	rep(i,N) chmax(ans, s[i]);

//	set<ll> st;
//	map<ll,ll> mp;
//	rep(i,N) st.insert(x[i]);
//	for(auto t: st) mp[t] = mp.size();

	RMQ<ll> sg(200005);
	ll sum = 0;
	sg.update(0,0);
	rep(i,N){
		sum += s[i];
		ll mn  = sg.get(0,i+1);
		chmax(ans, sum - mn);
		ll d = x[i+1] - x[i];
		sum -= d;
		sg.update(i+1, sum);
	}

	cout << ans << endl;

	return 0;
}

```