#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 = 1e18+7;


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];

	ll sum = 0, mn = 0;
	rep(i,N){
		sum += s[i];
		chmax(ans, sum - mn);
		ll d = x[i+1] - x[i];
		sum -= d;
		chmin(mn, sum);
	}

	cout << ans << endl;

	return 0;
}

### Prompt

Generate a Cpp solution 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>

#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 = 1e18+7;


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];

	ll sum = 0, mn = 0;
	rep(i,N){
		sum += s[i];
		chmax(ans, sum - mn);
		ll d = x[i+1] - x[i];
		sum -= d;
		chmin(mn, sum);
	}

	cout << ans << endl;

	return 0;
}

```