#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n;

    vector<int64_t> x(n), s(n + 1);
    for (int i = 0; i < n; i++) {
        cin >> x[i] >> s[i + 1];
        s[i + 1] += s[i];
    }

    vector<int> l(n), r(n);
    int i1;
    int64_t m;

    i1 = -1;
    for (int i = 0; i < n; i++) {
        int64_t t = -(s[i] - x[i]);
        if (i1 < 0 || t > m) {
            m = t;
            i1 = i;
        }
        l[i] = i1;
    }
    i1 = -1;
    for (int i = n - 1; i >= 0; i--) {
        int64_t t = s[i + 1] - x[i];
        if (i1 < 0 || t > m) {
            m = t;
            i1 = i;
        }
        r[i] = i1;
    }

    int64_t ans = 0;
    for (int i = 0; i < n; i++) {
        ans = max(ans, (s[r[i] + 1] - x[r[i]]) - (s[l[i]] - x[l[i]]));
    }
    cout << ans << endl;

    return 0;
}
### Prompt

Generate 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 <vector>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n;

    vector<int64_t> x(n), s(n + 1);
    for (int i = 0; i < n; i++) {
        cin >> x[i] >> s[i + 1];
        s[i + 1] += s[i];
    }

    vector<int> l(n), r(n);
    int i1;
    int64_t m;

    i1 = -1;
    for (int i = 0; i < n; i++) {
        int64_t t = -(s[i] - x[i]);
        if (i1 < 0 || t > m) {
            m = t;
            i1 = i;
        }
        l[i] = i1;
    }
    i1 = -1;
    for (int i = n - 1; i >= 0; i--) {
        int64_t t = s[i + 1] - x[i];
        if (i1 < 0 || t > m) {
            m = t;
            i1 = i;
        }
        r[i] = i1;
    }

    int64_t ans = 0;
    for (int i = 0; i < n; i++) {
        ans = max(ans, (s[r[i] + 1] - x[r[i]]) - (s[l[i]] - x[l[i]]));
    }
    cout << ans << endl;

    return 0;
}
```