#include "iostream"
#include "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "unordered_map"
#include "unordered_set"
#include "iomanip"
#include "cmath"
#include "random"
#include "bitset"
#include "cstdio"
#include "numeric"
#include "cassert"
#include "ctime"

using namespace std;

constexpr long long int MOD = 1000000007;
//constexpr int MOD = 1000000007;
//constexpr int MOD = 998244353;
//constexpr long long int MOD = 998244353;
constexpr long double EPS = 1e-8;

long long int N, M, K, L, R, H, W;
//int N, M, K, L, R, H, W;


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

	cin >> N;
	vector<long long int>p(N);
	vector<long long int>s(N);
	for (int i = 0; i < N; i++) {
		cin >> p[i] >> s[i];
	}
	long long int ans = s.back();
	long long int box = s.back() - p.back();
	for (int i = N - 2; i >= 0; i--) {
		box = max(box + s[i], s[i]-p[i]);
		ans = max(ans, box + p[i]);
	}
	cout << ans << endl;
}
### Prompt

Create a solution in cpp for 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 "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "unordered_map"
#include "unordered_set"
#include "iomanip"
#include "cmath"
#include "random"
#include "bitset"
#include "cstdio"
#include "numeric"
#include "cassert"
#include "ctime"

using namespace std;

constexpr long long int MOD = 1000000007;
//constexpr int MOD = 1000000007;
//constexpr int MOD = 998244353;
//constexpr long long int MOD = 998244353;
constexpr long double EPS = 1e-8;

long long int N, M, K, L, R, H, W;
//int N, M, K, L, R, H, W;


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

	cin >> N;
	vector<long long int>p(N);
	vector<long long int>s(N);
	for (int i = 0; i < N; i++) {
		cin >> p[i] >> s[i];
	}
	long long int ans = s.back();
	long long int box = s.back() - p.back();
	for (int i = N - 2; i >= 0; i--) {
		box = max(box + s[i], s[i]-p[i]);
		ans = max(ans, box + p[i]);
	}
	cout << ans << endl;
}
```