#include <algorithm>
#include <bitset>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <memory>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

// c++11
#include <array>
#include <tuple>
#include <unordered_map>
#include <unordered_set>

#define mp make_pair
#define mt make_tuple
#define rep(i, n) for (int i = 0; i < (n); i++)

using namespace std;

using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;

const int INF = 1 << 29;
const ll LL_INF = 1LL << 60;
const double EPS = 1e-9;
const ll MOD = 1000000007;

const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
int N;
vector<pair<ll, ll>> XS;
const int MAX_N = 200010;
ll dp[MAX_N];
int main()
{
    cin >> N;
    XS.resize(N);
    for (int i = 0; i < N; i++)
    {
        ll x, s;
        cin >> x >> s;
        XS[i] = mp(x, s);
    }
    ll sum = 0;
    ll L = XS[0].first;
    ll res = 0;
    for (int i = 0; i < N; i++)
    {
        dp[i] = XS[i].second - XS[i].first + (sum + L);
        res = max(res, XS[i].second);
        res = max(res, dp[i]);
        if (sum + L + XS[i].second < XS[i].first + XS[i].second)
        {
            sum = XS[i].second;
            L = XS[i].first;
        }
        else
        {
            sum += XS[i].second;
        }
    }
    cout << res << endl;

    return 0;
}
### Prompt

In Cpp, your task is to solve the following problem:

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 <algorithm>
#include <bitset>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <memory>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

// c++11
#include <array>
#include <tuple>
#include <unordered_map>
#include <unordered_set>

#define mp make_pair
#define mt make_tuple
#define rep(i, n) for (int i = 0; i < (n); i++)

using namespace std;

using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;

const int INF = 1 << 29;
const ll LL_INF = 1LL << 60;
const double EPS = 1e-9;
const ll MOD = 1000000007;

const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
int N;
vector<pair<ll, ll>> XS;
const int MAX_N = 200010;
ll dp[MAX_N];
int main()
{
    cin >> N;
    XS.resize(N);
    for (int i = 0; i < N; i++)
    {
        ll x, s;
        cin >> x >> s;
        XS[i] = mp(x, s);
    }
    ll sum = 0;
    ll L = XS[0].first;
    ll res = 0;
    for (int i = 0; i < N; i++)
    {
        dp[i] = XS[i].second - XS[i].first + (sum + L);
        res = max(res, XS[i].second);
        res = max(res, dp[i]);
        if (sum + L + XS[i].second < XS[i].first + XS[i].second)
        {
            sum = XS[i].second;
            L = XS[i].first;
        }
        else
        {
            sum += XS[i].second;
        }
    }
    cout << res << endl;

    return 0;
}
```