#include <iostream>
#include <cmath>
#include <functional>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <tuple>
using namespace std;
template<typename Type>
class RMQ {
private:
	int size_; std::vector<Type> dat;
public:
	RMQ() : size_(0), dat(std::vector<Type>()) {};
	RMQ(int size__) {
		for (size_ = 1; size_ < size__; ) size_ <<= 1;
		dat.resize(size_ << 1, std::numeric_limits<Type>::max());
	}
	template<class T>
	RMQ(T begin_, T end_) {
		int n = end_ - begin_;
		for (size_ = 1; size_ < n; size_ <<= 1); dat.resize(size_ << 1, std::numeric_limits<Type>::max());
		for (int i = 0; i < n; i++) dat[i + size_] = *(begin_ + i);
		for (int i = size_ - 1; i > 0; i--) dat[i] = std::min(dat[i << 1], dat[(i << 1) + 1]);
	}
	inline unsigned size() { return size_; }
	inline void update(int i, Type x) {
		i += size_; dat[i] = x;
		while (i > 1) {
			i >>= 1;
			dat[i] = std::min(dat[i << 1], dat[i << 1 | 1]);
		}
	}
	inline Type query(int l, int r) {
		Type ret = std::numeric_limits<Type>::max();
		l += size_; r += size_;
		while (l != r) {
			if (l & 1) ret = std::min(ret, dat[l++]); l >>= 1;
			if (r & 1) ret = std::min(ret, dat[--r]); r >>= 1;
		}
		return ret;
	}
};
long long n, a[200009], s[200009], dp[200009], maxn = 0;
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i] >> s[i]; if (i == 1)dp[i] = s[i]; else dp[i] = dp[i - 1] + (s[i] - (a[i] - a[i - 1]));
	}
	RMQ<long long>X(n + 1);
	for (int i = 1; i <= n; i++)X.update(i, -dp[i]);
	for (int i = 1; i <= n; i++) {
		maxn = max(maxn, (s[i] - dp[i]) - X.query(i, n + 1));
	}
	cout << maxn << endl;
	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 <iostream>
#include <cmath>
#include <functional>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <tuple>
using namespace std;
template<typename Type>
class RMQ {
private:
	int size_; std::vector<Type> dat;
public:
	RMQ() : size_(0), dat(std::vector<Type>()) {};
	RMQ(int size__) {
		for (size_ = 1; size_ < size__; ) size_ <<= 1;
		dat.resize(size_ << 1, std::numeric_limits<Type>::max());
	}
	template<class T>
	RMQ(T begin_, T end_) {
		int n = end_ - begin_;
		for (size_ = 1; size_ < n; size_ <<= 1); dat.resize(size_ << 1, std::numeric_limits<Type>::max());
		for (int i = 0; i < n; i++) dat[i + size_] = *(begin_ + i);
		for (int i = size_ - 1; i > 0; i--) dat[i] = std::min(dat[i << 1], dat[(i << 1) + 1]);
	}
	inline unsigned size() { return size_; }
	inline void update(int i, Type x) {
		i += size_; dat[i] = x;
		while (i > 1) {
			i >>= 1;
			dat[i] = std::min(dat[i << 1], dat[i << 1 | 1]);
		}
	}
	inline Type query(int l, int r) {
		Type ret = std::numeric_limits<Type>::max();
		l += size_; r += size_;
		while (l != r) {
			if (l & 1) ret = std::min(ret, dat[l++]); l >>= 1;
			if (r & 1) ret = std::min(ret, dat[--r]); r >>= 1;
		}
		return ret;
	}
};
long long n, a[200009], s[200009], dp[200009], maxn = 0;
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i] >> s[i]; if (i == 1)dp[i] = s[i]; else dp[i] = dp[i - 1] + (s[i] - (a[i] - a[i - 1]));
	}
	RMQ<long long>X(n + 1);
	for (int i = 1; i <= n; i++)X.update(i, -dp[i]);
	for (int i = 1; i <= n; i++) {
		maxn = max(maxn, (s[i] - dp[i]) - X.query(i, n + 1));
	}
	cout << maxn << endl;
	return 0;
}
```