#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

int N;
ll A[200010],X[200010];
ll mem[200010];
ll MAX[200010];
ll aa[200010],bb[200010];
ll ans = 0;

int main()
{
    scanf("%d",&N);
    for(int i = 1; i <= N; i++)
    {
        scanf("%lld%lld",X + i,A + i);
        A[i] += A[i - 1];
    }
    for(int i = 1; i <= N; i++)aa[i] = A[i] - X[i];
    for(int i = 1; i <= N; i++)bb[i] = A[i - 1] - X[i];

    ll mm = -1001001001001001001;
    for(int i = N; i > 0; i--)
    {
        mm = max(mm,aa[i]);
        MAX[i] = mm;
    }

    for(int i = 1; i <= N; i++)
    {
        ans = max(ans,MAX[i] - bb[i]);
    }
    printf("%lld\n",ans);
    return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 long long ll;

int N;
ll A[200010],X[200010];
ll mem[200010];
ll MAX[200010];
ll aa[200010],bb[200010];
ll ans = 0;

int main()
{
    scanf("%d",&N);
    for(int i = 1; i <= N; i++)
    {
        scanf("%lld%lld",X + i,A + i);
        A[i] += A[i - 1];
    }
    for(int i = 1; i <= N; i++)aa[i] = A[i] - X[i];
    for(int i = 1; i <= N; i++)bb[i] = A[i - 1] - X[i];

    ll mm = -1001001001001001001;
    for(int i = N; i > 0; i--)
    {
        mm = max(mm,aa[i]);
        MAX[i] = mm;
    }

    for(int i = 1; i <= N; i++)
    {
        ans = max(ans,MAX[i] - bb[i]);
    }
    printf("%lld\n",ans);
    return 0;
}

```