#include <iostream>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;
typedef long long ll;
int N;
ll x[200000], s[200000];
ll sum_l[200000], sum_r[200000];
ll sum[200000];

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(10) << fixed;
    cin >> N;
    for(int i = 0; i < N; i++){
        cin >> x[i] >> s[i];
        if(i >= 1){
            sum[i] = sum[i-1]+s[i]; 
        }else{
            sum[i] = s[i];
        }
        sum_l[i] = sum[i]-s[i]-x[i];
        sum_r[i] = sum[i]-x[i];
    }
    ll ans = 0;
    ll sum_r_max = -1e18;
    for(int i = N-1; i >= 0; i--){
        sum_r_max = max(sum_r[i], sum_r_max);
        ans = max(ans, sum_r_max-sum_l[i]);
    }
    cout << ans << endl;
}
### 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 <iostream>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;
typedef long long ll;
int N;
ll x[200000], s[200000];
ll sum_l[200000], sum_r[200000];
ll sum[200000];

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(10) << fixed;
    cin >> N;
    for(int i = 0; i < N; i++){
        cin >> x[i] >> s[i];
        if(i >= 1){
            sum[i] = sum[i-1]+s[i]; 
        }else{
            sum[i] = s[i];
        }
        sum_l[i] = sum[i]-s[i]-x[i];
        sum_r[i] = sum[i]-x[i];
    }
    ll ans = 0;
    ll sum_r_max = -1e18;
    for(int i = N-1; i >= 0; i--){
        sum_r_max = max(sum_r[i], sum_r_max);
        ans = max(ans, sum_r_max-sum_l[i]);
    }
    cout << ans << endl;
}
```