#include <bits/stdc++.h>
#define rep(i, n) for (lli i = 0; i < (n); i++)
#define rrep(i, n) for (lli i = (n)-1; i >= 0; i--)
using namespace std;
using lli = long long int;
using ll = __int128;

int main()
{
    int n;
    cin >> n;
    lli x[200005], s[200005];
    rep(i, n) cin >> x[i] >> s[i];
    ll cur = 0;
    ll sum = 0;
    ll ans = 0;
    rep(i, n)
    {
        sum += s[i];
        ans = max(ans, (ll)s[i]);
        ans = max(ans, sum - (ll)x[i] - cur);
        cur = min(cur, sum - (ll)x[i] - (ll)s[i]);
    }
    cout << (lli)ans << endl;
}

### 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 <bits/stdc++.h>
#define rep(i, n) for (lli i = 0; i < (n); i++)
#define rrep(i, n) for (lli i = (n)-1; i >= 0; i--)
using namespace std;
using lli = long long int;
using ll = __int128;

int main()
{
    int n;
    cin >> n;
    lli x[200005], s[200005];
    rep(i, n) cin >> x[i] >> s[i];
    ll cur = 0;
    ll sum = 0;
    ll ans = 0;
    rep(i, n)
    {
        sum += s[i];
        ans = max(ans, (ll)s[i]);
        ans = max(ans, sum - (ll)x[i] - cur);
        cur = min(cur, sum - (ll)x[i] - (ll)s[i]);
    }
    cout << (lli)ans << endl;
}

```