#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

const double EPS = 1e-10;
const ll INF = 100000000;
const ll MOD = 1000000007;

int n;
ll x[200000], s[200000];

int main() {
    cin >> n;
    rep(i,n) {
        cin >> x[i] >> s[i];
    }
    ll ans = 0;
    ll val = 0;
    rep(i,n) {
        if (i == 0) {
            ans = s[i];
            val = s[i];
            continue;
        }
        val -= x[i]-x[i-1];
        val = max(val, 0LL);
        val += s[i];
        ans = max(ans, val);
    }
    cout << ans << endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

const double EPS = 1e-10;
const ll INF = 100000000;
const ll MOD = 1000000007;

int n;
ll x[200000], s[200000];

int main() {
    cin >> n;
    rep(i,n) {
        cin >> x[i] >> s[i];
    }
    ll ans = 0;
    ll val = 0;
    rep(i,n) {
        if (i == 0) {
            ans = s[i];
            val = s[i];
            continue;
        }
        val -= x[i]-x[i-1];
        val = max(val, 0LL);
        val += s[i];
        ans = max(ans, val);
    }
    cout << ans << endl;
}

```