#include <bits/stdc++.h>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define svec(v) cout<<#v<<":";rep(kbrni,v.size())cout<<" "<<v[kbrni];cout<<endl
#define sset(s) cout<<#s<<":";each(kbrni,s)cout<<" "<<kbrni;cout<<endl
#define smap(m) cout<<#m<<":";each(kbrni,m)cout<<" {"<<kbrni.first<<":"<<kbrni.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 100005;

template<typename V> class segtree {
private:
    int n,sz;
    vector<V> node;
public:
    segtree(vector<V>& v){
        sz = (int)v.size();
        n = 1;
        while(n < sz){
            n *= 2;
        }
        node.resize(2*n-1);
        rep(i,sz){
            node[i+n-1] = v[i];
        }
        for(int i=n-2; i>=0; i--){
            node[i] = min(node[i*2+1],node[i*2+2]);
        }
    }
    void update(int k,V a)
    {
    	k += n-1;
    	node[k] = a;
    	while(k>0){
    		k = (k-1)/2;
    		node[k] = min(node[2*k+1],node[2*k+2]);
    	}
    }
    V query(int a,int b,int k=0,int l=0,int r=-1)
    {
        if(r < 0) r = n;
    	if(r <= a || b <= l){
    		return numeric_limits<V>::max();
    	}
    	if(a <= l && r <= b){
    		return node[k];
    	}else{
    		V vl = query(a,b,2*k+1,l,(l+r)/2);
    		V vr = query(a,b,2*k+2,(l+r)/2,r);
    		return min(vl,vr);
    	}
    }
    void print()
    {
        rep(i,sz){
            cout << query(i,i+1) << " ";
        }
        cout << endl;
    }
};

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vl x(n),s(n);
    vl sm(n+1);
    rep(i,n){
        cin >> x[i] >> s[i];
    }
    sm[0] = 0;
    rep(i,n){
        sm[i+1] = sm[i] + s[i];
    }
    vl dp(n),tmp(n);
    rep(i,n){
        tmp[i] = sm[i+1] - x[i];
    }
    rep(i,n){
        dp[i] = sm[i] - x[i];
    }
    segtree<ll> st(dp);
    ll mx = 0;
    rep(i,n){
        cmx(mx,tmp[i]-st.query(0,i+1));
    }
    cout << mx << "\n";
    return 0;
}

### Prompt

Please formulate 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>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define svec(v) cout<<#v<<":";rep(kbrni,v.size())cout<<" "<<v[kbrni];cout<<endl
#define sset(s) cout<<#s<<":";each(kbrni,s)cout<<" "<<kbrni;cout<<endl
#define smap(m) cout<<#m<<":";each(kbrni,m)cout<<" {"<<kbrni.first<<":"<<kbrni.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 100005;

template<typename V> class segtree {
private:
    int n,sz;
    vector<V> node;
public:
    segtree(vector<V>& v){
        sz = (int)v.size();
        n = 1;
        while(n < sz){
            n *= 2;
        }
        node.resize(2*n-1);
        rep(i,sz){
            node[i+n-1] = v[i];
        }
        for(int i=n-2; i>=0; i--){
            node[i] = min(node[i*2+1],node[i*2+2]);
        }
    }
    void update(int k,V a)
    {
    	k += n-1;
    	node[k] = a;
    	while(k>0){
    		k = (k-1)/2;
    		node[k] = min(node[2*k+1],node[2*k+2]);
    	}
    }
    V query(int a,int b,int k=0,int l=0,int r=-1)
    {
        if(r < 0) r = n;
    	if(r <= a || b <= l){
    		return numeric_limits<V>::max();
    	}
    	if(a <= l && r <= b){
    		return node[k];
    	}else{
    		V vl = query(a,b,2*k+1,l,(l+r)/2);
    		V vr = query(a,b,2*k+2,(l+r)/2,r);
    		return min(vl,vr);
    	}
    }
    void print()
    {
        rep(i,sz){
            cout << query(i,i+1) << " ";
        }
        cout << endl;
    }
};

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vl x(n),s(n);
    vl sm(n+1);
    rep(i,n){
        cin >> x[i] >> s[i];
    }
    sm[0] = 0;
    rep(i,n){
        sm[i+1] = sm[i] + s[i];
    }
    vl dp(n),tmp(n);
    rep(i,n){
        tmp[i] = sm[i+1] - x[i];
    }
    rep(i,n){
        dp[i] = sm[i] - x[i];
    }
    segtree<ll> st(dp);
    ll mx = 0;
    rep(i,n){
        cmx(mx,tmp[i]-st.query(0,i+1));
    }
    cout << mx << "\n";
    return 0;
}

```