#include <cstdio>
#include <vector>
#define repeat(i, n) for (int i = 0; (i) < int(n); ++(i))
using ll = long long;
using namespace std;
template <class T> inline void setmax(T & a, T const & b) { a = max(a, b); }

int main() {
    // input
    int n; scanf("%d", &n);
    vector<ll> x(n), s(n); repeat (i, n) scanf("%lld%lld", &x[i], &s[i]);
    // solve
    ll result = s[0];
    ll acc = s[0];
    repeat (i, n - 1) {
        acc -= x[i + 1] - x[i];
        setmax(acc, 0ll);
        acc += s[i + 1];
        setmax(result, acc);
    }
    // output
    printf("%lld\n", result);
    return 0;
}

### Prompt

Develop a solution in cpp 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 <cstdio>
#include <vector>
#define repeat(i, n) for (int i = 0; (i) < int(n); ++(i))
using ll = long long;
using namespace std;
template <class T> inline void setmax(T & a, T const & b) { a = max(a, b); }

int main() {
    // input
    int n; scanf("%d", &n);
    vector<ll> x(n), s(n); repeat (i, n) scanf("%lld%lld", &x[i], &s[i]);
    // solve
    ll result = s[0];
    ll acc = s[0];
    repeat (i, n - 1) {
        acc -= x[i + 1] - x[i];
        setmax(acc, 0ll);
        acc += s[i + 1];
        setmax(result, acc);
    }
    // output
    printf("%lld\n", result);
    return 0;
}

```