#include <bits/stdc++.h>

using namespace std;

using int64 = long long;

int main()
{
  int64 N, X[200000], S[200000];

  cin >> N;
  for(int i = 0; i < N; i++) {
    cin >> X[i] >> S[i];
  }

  // ∑_{i=L}^{R} S[i] - (X[R]-X[L])
  // ∑_{R} S[i] - ∑_{L-1} S[i] + (X[L] - X[R])
  // Z[i] = ∑_{i} S[i]
  // Z[R]-Z[L-1]+X[L]-X[R]
  // (X[L]-Z[L-1]) + (Z[R]-X[R])

  int64 ret = 0;
  int64 beet = 0, sum = 0;
  for(int i = 0; i < N; i++) {
    beet = max(beet, X[i] - sum);
    sum += S[i];
    ret = max(ret, beet + sum - X[i]);
  }
  cout << ret << endl;
}
### 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;

using int64 = long long;

int main()
{
  int64 N, X[200000], S[200000];

  cin >> N;
  for(int i = 0; i < N; i++) {
    cin >> X[i] >> S[i];
  }

  // ∑_{i=L}^{R} S[i] - (X[R]-X[L])
  // ∑_{R} S[i] - ∑_{L-1} S[i] + (X[L] - X[R])
  // Z[i] = ∑_{i} S[i]
  // Z[R]-Z[L-1]+X[L]-X[R]
  // (X[L]-Z[L-1]) + (Z[R]-X[R])

  int64 ret = 0;
  int64 beet = 0, sum = 0;
  for(int i = 0; i < N; i++) {
    beet = max(beet, X[i] - sum);
    sum += S[i];
    ret = max(ret, beet + sum - X[i]);
  }
  cout << ret << endl;
}
```