Problem:
Score : 200 points

Problem Statement

In a factory, there are N robots placed on a number line.
Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative.

We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect.
Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints.

Find the maximum number of robots that we can keep.

Constraints

1 \leq N \leq 100,000

0 \leq X_i \leq 10^9 (1 \leq i \leq N)

1 \leq L_i \leq 10^9 (1 \leq i \leq N)

If i \neq j, X_i \neq X_j.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 L_1
X_2 L_2
\vdots
X_N L_N

Output

Print the maximum number of robots that we can keep.

Sample Input 1

4
2 4
4 3
9 3
100 5

Sample Output 1

3

By removing Robot 2, we can keep the other three robots.

Sample Input 2

2
8 20
1 10

Sample Output 2

1

Sample Input 3

5
10 1
2 1
4 1
6 1
8 1

Sample Output 3

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include<limits.h>
#include<stdio.h>
#include<cstring>
#include<string>
using namespace  std;
typedef long long ll;
typedef long double ld;
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pb push_back
#define mp make_pair
#define sz(a) a.size()
#define fi first
#define se second
#define rsz(a,n) a.resize(n)
#define test(t) ll t;cin>>t;while(t--)
#define forn(i,e) for(int i = 0; i < e; i++)
#define forsn(i,s,e) for(int i = s; i < e; i++)
#define rforn(i,s) for(int i = s; i >= 0; i--)
#define rforsn(i,s,e) for(int i = s; i >= e; i--)
#define fillv(a,k) memset(a,k,sizeof(a))
#define leadzero(a) __builtin_clz(a) //count leading zeros
#define trailzero(a) __builtin_ctz(a) //count trailing zeros
#define bitcount(a) __builtin_popcount(a) // count set bits (add ll)
#define ln cout<<endl
#define sp cout<<" "
#define spaceprint(a,i,s,n) {forsn(i,s,n) { cout<<a[i]; sp;}}
#define lineprint(a,i,s,n) {forsn(i,s,n) {cout<<a[i]; ln;}}
#define maxe(a) *max_element(a.begin(),a.end())
#define maxi(a) max_element(a.begin(),a.end())-a.begin()
#define mine(a) *min_element(a.begin(),a.end())
#define mini(a) min_element(a.begin(),a.end())-a.begin()
#define all(c) c.begin(),c.end()
#define trav(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define present(container, element) (container.find(element) != container.end())
#define cpresent(container, element)  (find(all(container),element) != container.end())// check the presence of element
//copy(from_begin, from_end, to_begin); copy function
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<ll,ll> p64;
typedef pair<int,int> p32;
typedef pair<int,p32> p96;
typedef vector<ll> v64;
typedef vector<string> vs;
typedef vector<int> v32;
typedef vector<v32> vv32;
typedef vector<v64> vv64;
typedef vector<p32> vp32;
typedef vector<p64> vp64;
typedef vector<vp32> vvp32;
typedef map<int,int> m32;
typedef map<ll,ll> m64;
const int LIM = 1e5+5, MOD =  1e9+7;
#define sumv(v) accumulate(all(v),ll(0))
#define productv(v) accumulate(all(v), ll(1), multiplies< ll >())
ll gcd(ll a, ll b) { if(b == 0) return a; return gcd(b, a % b); }
ll fastpowMOD(ll a, ll p,ll MOD){ if(p==0) return 1; ll z = fastpowMOD(a,p/2,MOD); z = (z*z)%MOD; if(p%2) z = (z*a)%MOD; return z; }
bool seive[100005];
ll dp[10005][105];
void SieveOfEratosthenes(ll n)
{ memset(seive, true, sizeof(seive)); for (ll p=2; p*p<=n; p++) if (seive[p] == true) for (ll i=p*p; i<=n; i += p) seive[i] = false; }
p64 a[100005];
int main()
{
    fast;
    ll n,i,j;
    cin>>n;
    for(i=0;i<n;i++)
    {
        ll x,l;
        cin>>x>>l;
        a[i].fi=x-l;
        a[i].se=x+l;
    }
    ll k=0;
    //ll j;
    i=0;
    j=0;
    sort(a,a+n);
    while(i<n)
    {
        for(j=i+1;j<n;j++)
        {
            if(a[j].fi>a[i].se)
            {
                i=j;
                break;
            }
            else
            {
                if(a[j].se<=a[i].se)
                {
                    k++;
                    i=j;
                    break;
                }
                else
                {
                    k++;
                    continue;
                }
            }
        }
        if(j==n)
            break;
    }
    cout<<n-k;

}