//#include <bits/stdc++.h>
 
#include <iostream>
#include <algorithm>
 
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <cstring>
#include <string>
#include <utility>
#include <array>
#include <complex>
#include <valarray>
 
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <chrono>
#include <random>
#include <numeric>
 
 
using namespace std;
#define int long long
 
typedef long long ll;
typedef unsigned long long ull;
//typedef unsigned __int128 HASH;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pullull;
typedef pair<ll,int> plli;
typedef pair<double, int> pdbi;
typedef pair<int,pii> pipii;
typedef pair<ll,pll> plpll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<pii> vpii;
typedef vector<vector<int>> mat;
 
#define rep(i,n) for (int i=0;i<(n);i++)
#define rep2(i,a,b) for (int i=(a);i<(b);i++)
#define rrep(i,n) for (int i=(n);i>0;i--)
#define rrep2(i,a,b) for (int i=(a);i>b;i--)
#define pb push_back
#define fi first
#define se second
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
 
const ll hmod1 = 999999937;
const ll hmod2 = 1000000000 + 9;
const int INF = 1<<30;
const ll INFLL = 1LL<<62;
const double EPS = 1e-12;
const ll mod = 1000000000 + 7;
const int dx4[4] = {1, 0, -1, 0};
const int dy4[4] = {0, 1, 0, -1};
const int dx8[8] = {1, 1, 1, 0, 0, -1, -1, -1};
const int dy8[8] = {0, 1, -1, 1, -1, 0, 1, -1};
const double pi = 3.141592653589793;
 
#define addm(X, Y) (X) = ((X) + ((Y) % mod) + mod) % mod
#define inside(y, x, h, w) (0 <= (y) && (y) < (h) && 0 <= (x) && (x) < (w)) ? true : false

int n;
int x[200000 + 5], s[200000 + 5];

struct segtree {
    int size;
    vector<long long> node;

    segtree (int n) : size(1) {
        while (size < n) size *= 2;
        node.resize(2 * size - 1, -INFLL / 10);
    }

    void update(int k, int a) {
        k += size - 1;
        node[k] = a;
        while (k > 0){
            k = (k - 1) / 2;
            node[k] = max(node[2 * k + 1], node[2 * k + 2]);
        }
    }
    //[queryL: queryR)
    long long get(int queryL, int queryR, int k = 0, int nodeL = 0, int nodeR = -1) {
        if (nodeR == -1) nodeR = size;
        if (nodeR <= queryL || queryR <= nodeL) return -INFLL / 10;
        if (queryL <= nodeL && nodeR <= queryR) return node[k];
        else{
            int nodeM = (nodeL + nodeR) / 2;
            long long vl = get(queryL, queryR, 2 * k + 1, nodeL, nodeM);
            long long vr = get(queryL, queryR, 2 * k + 2, nodeM, nodeR);
            return max(vl, vr);
        }
    }
};

int sum[200000 + 5];

signed main() {	
    cin.tie(0);
    ios::sync_with_stdio(false);
	cin >> n;
	segtree seg(n);
	rep(i, n) {
		cin >> x[i] >> s[i];
		if (i == 0) {
			seg.update(i, s[i]);
			sum[i] += s[i];
		}
		else {
			sum[i] += sum[i - 1] + s[i] - (x[i] - x[i - 1]);
			seg.update(i, sum[i]);
		}
	}
	int ans = 0;
	rep(i, n) {
		if (i == 0) ans = max(ans, seg.get(0, n));
		else ans = max(ans, seg.get(i, n) - sum[i - 1] + (x[i] - x[i - 1]));
	} 
	cout << ans << endl;
}

### Prompt

Your task is to create a CPP solution to 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 <bits/stdc++.h>
 
#include <iostream>
#include <algorithm>
 
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <cstring>
#include <string>
#include <utility>
#include <array>
#include <complex>
#include <valarray>
 
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <chrono>
#include <random>
#include <numeric>
 
 
using namespace std;
#define int long long
 
typedef long long ll;
typedef unsigned long long ull;
//typedef unsigned __int128 HASH;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pullull;
typedef pair<ll,int> plli;
typedef pair<double, int> pdbi;
typedef pair<int,pii> pipii;
typedef pair<ll,pll> plpll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<pii> vpii;
typedef vector<vector<int>> mat;
 
#define rep(i,n) for (int i=0;i<(n);i++)
#define rep2(i,a,b) for (int i=(a);i<(b);i++)
#define rrep(i,n) for (int i=(n);i>0;i--)
#define rrep2(i,a,b) for (int i=(a);i>b;i--)
#define pb push_back
#define fi first
#define se second
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
 
const ll hmod1 = 999999937;
const ll hmod2 = 1000000000 + 9;
const int INF = 1<<30;
const ll INFLL = 1LL<<62;
const double EPS = 1e-12;
const ll mod = 1000000000 + 7;
const int dx4[4] = {1, 0, -1, 0};
const int dy4[4] = {0, 1, 0, -1};
const int dx8[8] = {1, 1, 1, 0, 0, -1, -1, -1};
const int dy8[8] = {0, 1, -1, 1, -1, 0, 1, -1};
const double pi = 3.141592653589793;
 
#define addm(X, Y) (X) = ((X) + ((Y) % mod) + mod) % mod
#define inside(y, x, h, w) (0 <= (y) && (y) < (h) && 0 <= (x) && (x) < (w)) ? true : false

int n;
int x[200000 + 5], s[200000 + 5];

struct segtree {
    int size;
    vector<long long> node;

    segtree (int n) : size(1) {
        while (size < n) size *= 2;
        node.resize(2 * size - 1, -INFLL / 10);
    }

    void update(int k, int a) {
        k += size - 1;
        node[k] = a;
        while (k > 0){
            k = (k - 1) / 2;
            node[k] = max(node[2 * k + 1], node[2 * k + 2]);
        }
    }
    //[queryL: queryR)
    long long get(int queryL, int queryR, int k = 0, int nodeL = 0, int nodeR = -1) {
        if (nodeR == -1) nodeR = size;
        if (nodeR <= queryL || queryR <= nodeL) return -INFLL / 10;
        if (queryL <= nodeL && nodeR <= queryR) return node[k];
        else{
            int nodeM = (nodeL + nodeR) / 2;
            long long vl = get(queryL, queryR, 2 * k + 1, nodeL, nodeM);
            long long vr = get(queryL, queryR, 2 * k + 2, nodeM, nodeR);
            return max(vl, vr);
        }
    }
};

int sum[200000 + 5];

signed main() {	
    cin.tie(0);
    ios::sync_with_stdio(false);
	cin >> n;
	segtree seg(n);
	rep(i, n) {
		cin >> x[i] >> s[i];
		if (i == 0) {
			seg.update(i, s[i]);
			sum[i] += s[i];
		}
		else {
			sum[i] += sum[i - 1] + s[i] - (x[i] - x[i - 1]);
			seg.update(i, sum[i]);
		}
	}
	int ans = 0;
	rep(i, n) {
		if (i == 0) ans = max(ans, seg.get(0, n));
		else ans = max(ans, seg.get(i, n) - sum[i - 1] + (x[i] - x[i - 1]));
	} 
	cout << ans << endl;
}

```