#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 5;
int n;
ll x[maxn], s[maxn];
ll dp[maxn];
int main(){
//    freopen("e:\\in.txt","r",stdin);
    while(cin>>n){
        for(int i = 0; i < n; i++)cin>>x[i]>>s[i];
        dp[0] = s[0];
        ll ans = dp[0];
        for(int i = 1; i < n; i++){
            dp[i] = max(dp[i - 1] + s[i] - x[i] + x[i - 1], s[i]);
            ans = max(ans, dp[i]);
        }
        cout<<ans<<endl;
    }
    return 0;
}

### Prompt

Please formulate 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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 5;
int n;
ll x[maxn], s[maxn];
ll dp[maxn];
int main(){
//    freopen("e:\\in.txt","r",stdin);
    while(cin>>n){
        for(int i = 0; i < n; i++)cin>>x[i]>>s[i];
        dp[0] = s[0];
        ll ans = dp[0];
        for(int i = 1; i < n; i++){
            dp[i] = max(dp[i - 1] + s[i] - x[i] + x[i - 1], s[i]);
            ans = max(ans, dp[i]);
        }
        cout<<ans<<endl;
    }
    return 0;
}

```