#include <bits/stdc++.h>
using namespace std;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define Ans(f, y, n) if(f) cout << y << endl; else cout << n << endl;
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) cout << #v << "=" << endl; REP(i, v.size()){ cout << v[i] << ","; } cout << endl;
#define mdebug(m) cout << #m << "=" << endl; REP(i, m.size()){ REP(j, m[i].size()){ cout << m[i][j] << ","; } cout << endl;}
#define pb push_back
#define f first
#define s second
#define int long long
#define INF 1000000000000000000

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;

template<typename T> void readv(vector<T> &a){ REP(i, a.size()) cin >> a[i]; }
void readv_m1(vector<int> &a){ REP(i, a.size()){cin >> a[i]; a[i]--;} }

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};

int mod = 1000000007;
//int mod = 998244353;
#define Add(x, y) x = (x + (y)) % mod
#define Mult(x, y) x = (x * (y)) % mod
#define AddMult(x, y, z) x = (x + (y) * (z)) % mod

class Segment_tree
{
    using data_type = int;

public:

    vector<data_type> dat;
    int N;
    data_type id;

    data_type func(data_type u, data_type v){
        return max(u, v);
    }

    Segment_tree(int n, data_type id): id(id) {
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<data_type>(2 * N - 1, id);
    }

    Segment_tree(int n, data_type id, vector<data_type> v): id(id) {
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<data_type>(2 * N - 1, id);
        REP(i, n) dat[i + N - 1] = v[i];
        IREP(i, N - 1) dat[i] = func(dat[i * 2 + 1], dat[i * 2 + 2]);
    }

    //k番目の値をaに更新
    void update(int k, data_type a){
        k += N - 1;
        dat[k] = a;
        while(k > 0){
            k = (k - 1) / 2;
            dat[k] = func(dat[k * 2 + 1], dat[k * 2 + 2]);
        }
    }

    data_type query(int a, int b, int k, int l, int r){
        if(r <= a || b <= l) return id;
        if(a <= l && r <= b) return dat[k];
        else{
            data_type vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
            data_type vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
            return func(vl, vr);
        }
    }

    //[a, b)の演算結果
    data_type query(int a, int b){
        return query(a, b, 0, 0, N);
    }

    data_type getdat(int k){
        return dat[k + N - 1];
    }
};

signed main(){

    int N; cin >> N;
    vec x(N), s(N);
    REP(i, N){
        cin >> x[i] >> s[i];
    }
    vec v(N);
    int t = 0;
    REP(i, N){
        t += s[i];
        v[i] = t - x[i];
    }
    Segment_tree ST(N, -INF, v);
    int ans = -INF;
    t = 0;
    REP(i, N){
        int val = ST.query(i, N);
        val = val + x[i] - t;
        Max(ans, val);
        t += s[i];
    }
    cout << ans << endl;

    return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define Ans(f, y, n) if(f) cout << y << endl; else cout << n << endl;
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) cout << #v << "=" << endl; REP(i, v.size()){ cout << v[i] << ","; } cout << endl;
#define mdebug(m) cout << #m << "=" << endl; REP(i, m.size()){ REP(j, m[i].size()){ cout << m[i][j] << ","; } cout << endl;}
#define pb push_back
#define f first
#define s second
#define int long long
#define INF 1000000000000000000

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;

template<typename T> void readv(vector<T> &a){ REP(i, a.size()) cin >> a[i]; }
void readv_m1(vector<int> &a){ REP(i, a.size()){cin >> a[i]; a[i]--;} }

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};

int mod = 1000000007;
//int mod = 998244353;
#define Add(x, y) x = (x + (y)) % mod
#define Mult(x, y) x = (x * (y)) % mod
#define AddMult(x, y, z) x = (x + (y) * (z)) % mod

class Segment_tree
{
    using data_type = int;

public:

    vector<data_type> dat;
    int N;
    data_type id;

    data_type func(data_type u, data_type v){
        return max(u, v);
    }

    Segment_tree(int n, data_type id): id(id) {
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<data_type>(2 * N - 1, id);
    }

    Segment_tree(int n, data_type id, vector<data_type> v): id(id) {
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<data_type>(2 * N - 1, id);
        REP(i, n) dat[i + N - 1] = v[i];
        IREP(i, N - 1) dat[i] = func(dat[i * 2 + 1], dat[i * 2 + 2]);
    }

    //k番目の値をaに更新
    void update(int k, data_type a){
        k += N - 1;
        dat[k] = a;
        while(k > 0){
            k = (k - 1) / 2;
            dat[k] = func(dat[k * 2 + 1], dat[k * 2 + 2]);
        }
    }

    data_type query(int a, int b, int k, int l, int r){
        if(r <= a || b <= l) return id;
        if(a <= l && r <= b) return dat[k];
        else{
            data_type vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
            data_type vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
            return func(vl, vr);
        }
    }

    //[a, b)の演算結果
    data_type query(int a, int b){
        return query(a, b, 0, 0, N);
    }

    data_type getdat(int k){
        return dat[k + N - 1];
    }
};

signed main(){

    int N; cin >> N;
    vec x(N), s(N);
    REP(i, N){
        cin >> x[i] >> s[i];
    }
    vec v(N);
    int t = 0;
    REP(i, N){
        t += s[i];
        v[i] = t - x[i];
    }
    Segment_tree ST(N, -INF, v);
    int ans = -INF;
    t = 0;
    REP(i, N){
        int val = ST.query(i, N);
        val = val + x[i] - t;
        Max(ans, val);
        t += s[i];
    }
    cout << ans << endl;

    return 0;
}

```