#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;

int main() {
        int n;
        scanf("%d", &n);
        vector<long long> all;
        vector<long long> x(n);
        for (int i = 0; i < n; i ++) {
                long long s;
                scanf("%lld%lld", &x[i], &s);
                if (i == 0) all.push_back(s);
                else {
                        all.push_back(-(x[i] - x[i - 1]));
                        all.push_back(s);
                }
        }
        n = all.size();
        for (int i = 1; i < n; i ++) all[i] += all[i - 1];
        vector<long long> ma(n); //max of [l, n)
        ma[n - 1] = all[n - 1];
        for (int i = n - 2; i >= 0; i --) ma[i] = max(ma[i + 1], all[i]);
        long long ans = 0;
        ans = max(ans, ma[0]);
        for (int l = 0; l < n - 1; l ++) ans = max(ans, ma[l + 1] - all[l]);
        printf("%lld\n", ans);
        return 0;
}


### Prompt

Your challenge is to write 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 <cstdio>
#include <algorithm>
#include <vector>
using namespace std;

int main() {
        int n;
        scanf("%d", &n);
        vector<long long> all;
        vector<long long> x(n);
        for (int i = 0; i < n; i ++) {
                long long s;
                scanf("%lld%lld", &x[i], &s);
                if (i == 0) all.push_back(s);
                else {
                        all.push_back(-(x[i] - x[i - 1]));
                        all.push_back(s);
                }
        }
        n = all.size();
        for (int i = 1; i < n; i ++) all[i] += all[i - 1];
        vector<long long> ma(n); //max of [l, n)
        ma[n - 1] = all[n - 1];
        for (int i = n - 2; i >= 0; i --) ma[i] = max(ma[i + 1], all[i]);
        long long ans = 0;
        ans = max(ans, ma[0]);
        for (int l = 0; l < n - 1; l ++) ans = max(ans, ma[l + 1] - all[l]);
        printf("%lld\n", ans);
        return 0;
}


```