#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int N = 2e5 + 7;
const int SS = 4 * N;
const ll INF = 1e17 + 7;
int n;

ll x[N], s[N];
ll lazy[SS];
ll seg[SS];

void update(int v, ll val)
{
    lazy[v] += val;
    seg[v] += val;
    return;
}

void shift(int v)
{
    update(2 * v, lazy[v]);
    update(2 * v + 1, lazy[v]);
    lazy[v] = 0;
}

void add(int l, int r, ll val, int s = 0, int e = n, int v = 1)
{
    if (l >= e || r <= s)
        return;
    if (l <= s && e <= r)
    {
        update(v, val);
        return;
    }
    int mid = (s + e) / 2;
    shift(v);
    add(l, r, val, s, mid, 2 * v);
    add(l, r, val, mid, e, 2 * v + 1);
    seg[v] = max(seg[2 * v], seg[2 * v + 1]);
}


int main()
{
    ios::sync_with_stdio(false);

    cin >> n;
    for (int i=0; i<n; i++)
        cin >> x[i] >> s[i];

    for (int i=0; i<n; i++)
        add(i, i+1, -x[i]);
    for (int i=0; i<n; i++)
        add(i, n, s[i]);
    ll ans = 0;
    for (int i=0; i<n; i++)
    {
        ans = max(ans, x[i] + seg[1]);
        add(i, n, -s[i]);
        add(i, i+1, -INF);
    }
    cout << ans << "\n";
    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 <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int N = 2e5 + 7;
const int SS = 4 * N;
const ll INF = 1e17 + 7;
int n;

ll x[N], s[N];
ll lazy[SS];
ll seg[SS];

void update(int v, ll val)
{
    lazy[v] += val;
    seg[v] += val;
    return;
}

void shift(int v)
{
    update(2 * v, lazy[v]);
    update(2 * v + 1, lazy[v]);
    lazy[v] = 0;
}

void add(int l, int r, ll val, int s = 0, int e = n, int v = 1)
{
    if (l >= e || r <= s)
        return;
    if (l <= s && e <= r)
    {
        update(v, val);
        return;
    }
    int mid = (s + e) / 2;
    shift(v);
    add(l, r, val, s, mid, 2 * v);
    add(l, r, val, mid, e, 2 * v + 1);
    seg[v] = max(seg[2 * v], seg[2 * v + 1]);
}


int main()
{
    ios::sync_with_stdio(false);

    cin >> n;
    for (int i=0; i<n; i++)
        cin >> x[i] >> s[i];

    for (int i=0; i<n; i++)
        add(i, i+1, -x[i]);
    for (int i=0; i<n; i++)
        add(i, n, s[i]);
    ll ans = 0;
    for (int i=0; i<n; i++)
    {
        ans = max(ans, x[i] + seg[1]);
        add(i, n, -s[i]);
        add(i, i+1, -INF);
    }
    cout << ans << "\n";
    return 0;
}

```