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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, start, end) for (long long i = start; i < end; ++i)
#define repreverse(i, start, end) for (long long i = start; i >= end; --i)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define len(x) ((long long)(x).size())
#define lcm(a, b) ((a) / __gcd((a), (b)) * (b))
using namespace std;
using ll = long long;
using ld = long double;
using vll = vector<ll>;
using vllvll = vector<vll>;
using vc = vector<char>;
using pll = pair<ll, ll>;
template<class T> void print1d(T x,ll n=-1){if(n==-1)n=x.size();rep(i,0,n){cout<<x[i]<<' ';}cout<<'\n';}
template<class T> void print2d(T x,ll r=-1,ll c=-1){if(r==-1)r=x.size();if(c==-1)c=x[0].size();rep(i,0,r)print1d(x[i],c);}
template<class T, class U> bool haskey(T mp, U key) { return mp.find(key) != mp.end(); }
template<class T, class U> bool isin(T el, U container) { return find(all(container), el) != container.end(); }
template<class T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
template<class T> bool even(T n) { return ! (n & 1); }
template<class T> bool odd(T n) { return n & 1; }
template<class T> ll rup(T a, T b) { return a % b ? a / b + 1 : a / b; }
template<class T> ld deg2rad(T deg) { return M_PI * deg / 180.0; }
template<class T> ld rad2deg(T rad) { return 180.0 * rad / M_PI; }
const long double pi = M_PI;
const long long big = 1LL << 50;
const long long inf = 1LL << 60;
const long long mod = 1e9 + 7;
ll intpow(ll a, ll n, ll _mod=numeric_limits<ll>::max()) { ll p=1; while (n) { if (n&1) p=p*a%_mod; a=a*a%_mod; n>>=1; } return p; } 
ll modc(ll a, char op, ll b, ll _mod=mod) { a %= _mod; b %= _mod; ll res = 1; switch(op) {
        case '+': res = (a + b) % _mod; break;
        case '-': res = (a - b) % _mod; break;
        case '*': res = a * b % _mod; break;
        case '/': res = modc(a, '*', modc(b, '^', _mod-2, _mod), _mod); break;
        case '^': res = intpow(a, b, _mod); break;
        case 'P': rep(i, a-b+1, a+1) res = modc(res, '*', i, _mod); break;
        case 'C': res = modc(modc(a, 'P', b, _mod), '/', modc(b, 'P', b, _mod)); break; } if (res < 0) res += _mod; return res; }


int main()
{
        ll N;
        cin >> N;
        vll left(N), right(N);

        vector<pll> v(N);
        rep(i, 0, N) {
                ll x, l;
                cin >> x >> l;
                v[i].first = x - l;
                v[i].second = x + l;
        }

        sort(all(v));

        ll tmp = -inf;
        ll ans = 0;
        rep(i, 0, N) {
                if (v[i].first >= tmp) {
                        ++ans;
                        tmp = v[i].second;
                }
                
        }

        cout << ans << endl;
}