#include <stdio.h>
#include <algorithm>
#include <assert.h>
#include <bitset>
#include <cmath>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits.h>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <time.h>
#include <unordered_map>
#include <unordered_set>
#include <vector>

#pragma warning(disable:4996)
#pragma comment(linker, "/STACK:336777216")
using namespace std;

#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define ldb ldouble

typedef tuple<int, int, int> t3;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <db, db> pdd;

int IT_MAX = 1 << 18;
int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const db PI = acos(-1);
const db ERR = 1e-10;
#define szz(x) (int)(x).size()
#define rep(i, n) for(int i=0;i<n;i++)
#define Se second
#define Fi first

const int MX = 200005;
const int MM = 1000000007;

ll A[MX], B[MX];
int N;

ll DnC(int s, int e){
	if( s > e ) return 0;
	if( s == e ) return B[s];
	int m = (s+e) / 2;
	ll l = 0, total = 0, r = 0;
	for(int i = m; i >= s; i--){
		total += B[i];
		l = max(l, total - A[m] + A[i]);
	}
	total = 0;
	for(int i = m; i <= e; i++){
		total += B[i];
		r = max(r, total - A[i] + A[m]);
	}
	return max({DnC(s, m-1), DnC(m+1, e), l+r-B[m]});
}

int main()
{
	scanf("%d", &N);
	for(int i = 1; i <= N; i++) scanf("%lld%lld", A+i, B+i);
	printf("%lld\n", DnC(1, N));
}

### Prompt

Please create a solution in CPP 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 <stdio.h>
#include <algorithm>
#include <assert.h>
#include <bitset>
#include <cmath>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits.h>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <time.h>
#include <unordered_map>
#include <unordered_set>
#include <vector>

#pragma warning(disable:4996)
#pragma comment(linker, "/STACK:336777216")
using namespace std;

#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define ldb ldouble

typedef tuple<int, int, int> t3;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <db, db> pdd;

int IT_MAX = 1 << 18;
int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const db PI = acos(-1);
const db ERR = 1e-10;
#define szz(x) (int)(x).size()
#define rep(i, n) for(int i=0;i<n;i++)
#define Se second
#define Fi first

const int MX = 200005;
const int MM = 1000000007;

ll A[MX], B[MX];
int N;

ll DnC(int s, int e){
	if( s > e ) return 0;
	if( s == e ) return B[s];
	int m = (s+e) / 2;
	ll l = 0, total = 0, r = 0;
	for(int i = m; i >= s; i--){
		total += B[i];
		l = max(l, total - A[m] + A[i]);
	}
	total = 0;
	for(int i = m; i <= e; i++){
		total += B[i];
		r = max(r, total - A[i] + A[m]);
	}
	return max({DnC(s, m-1), DnC(m+1, e), l+r-B[m]});
}

int main()
{
	scanf("%d", &N);
	for(int i = 1; i <= N; i++) scanf("%lld%lld", A+i, B+i);
	printf("%lld\n", DnC(1, N));
}

```