#include<bits/stdc++.h>
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<LL,LL> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
const LL mod=1000000007;
const LL LINF=1LL<<62;
const int INF=1<<30;
int dx[]={1,0,-1,0,1,-1,1,-1};
int dy[]={0,1,0,-1,1,-1,-1,1};



int main(){
    int n;cin >> n;
    vector<LL> x(n),s(n);
    for (int i = 0; i < n; i++) {
        cin >> x[i] >> s[i];
    }
    vector<LL> L(n,0),R(n,0);
    LL S = 0;
    for (int i = n-2; i >= 0; i--) {
        S = max(0LL, S + s[i+1] - (x[i+1] - x[i]));
        R[i] = S;
    }
    S = 0;
    for (int i = 1; i < n; i++) {
        S = max(0LL, S + s[i-1] - (x[i] - x[i-1])); 
        L[i] = S;
    }
    LL ans = 0;
    for (int i = 0; i < n; i++) {
        chmax(ans, L[i] + R[i] + s[i]);
    }
    cout << ans << endl;
    return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<LL,LL> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
const LL mod=1000000007;
const LL LINF=1LL<<62;
const int INF=1<<30;
int dx[]={1,0,-1,0,1,-1,1,-1};
int dy[]={0,1,0,-1,1,-1,-1,1};



int main(){
    int n;cin >> n;
    vector<LL> x(n),s(n);
    for (int i = 0; i < n; i++) {
        cin >> x[i] >> s[i];
    }
    vector<LL> L(n,0),R(n,0);
    LL S = 0;
    for (int i = n-2; i >= 0; i--) {
        S = max(0LL, S + s[i+1] - (x[i+1] - x[i]));
        R[i] = S;
    }
    S = 0;
    for (int i = 1; i < n; i++) {
        S = max(0LL, S + s[i-1] - (x[i] - x[i-1])); 
        L[i] = S;
    }
    LL ans = 0;
    for (int i = 0; i < n; i++) {
        chmax(ans, L[i] + R[i] + s[i]);
    }
    cout << ans << endl;
    return 0;
}

```