#include<bits/stdc++.h>
using namespace std;
typedef pair<long long,int>P;
const int maxn=2e5+5;
long long n,x[maxn],s[maxn],m;
signed main(){
    priority_queue<P>q;
    scanf("%lld",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld%lld",&x[i],&s[i]);
        s[i]+=s[i-1];q.push(P(s[i]-x[i],i));}
    for(int i=1,flag;i<=n;i++){
        flag=1;while(flag){
            P p=q.top();
            if(p.second<i)q.pop();
            else{m=max(m,p.first-s[i-1]+x[i]);flag=0;}
        }
    }
    printf("%lld\n",m); 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>
using namespace std;
typedef pair<long long,int>P;
const int maxn=2e5+5;
long long n,x[maxn],s[maxn],m;
signed main(){
    priority_queue<P>q;
    scanf("%lld",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld%lld",&x[i],&s[i]);
        s[i]+=s[i-1];q.push(P(s[i]-x[i],i));}
    for(int i=1,flag;i<=n;i++){
        flag=1;while(flag){
            P p=q.top();
            if(p.second<i)q.pop();
            else{m=max(m,p.first-s[i-1]+x[i]);flag=0;}
        }
    }
    printf("%lld\n",m); return 0;
}
```