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 <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <stdlib.h>
#include <string.h>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <stdio.h>
#include <tuple>
#include <bitset>
#include <cfloat>
#include <fstream>
#include <limits.h>
#include <list>
#include <math.h>
#include <random>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <numeric>
#include <utility>
#include <sstream>
#include <cassert>
template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
using namespace std;
#define int long long
#define ll long long
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define P pair<ll, ll>
#define sz(x) (ll)x.size()
#define ALL(x) (x).begin(),(x).end()
#define ALLR(x) (x).rbegin(),(x).rend()
#define VE vector<ll>
#define COUT(x) cout<<(x)<<endl
#define MA map<ll,ll>
#define SE set<ll>
#define PQ priority_queue<ll>
#define PQR priority_queue<ll,VE,greater<ll>>
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define EPS (1e-14)
#define pb push_back
long long MOD = 1000000007;
//const long long MOD = 998244353;
const long long INF = 1LL << 60;
const double PI = acos(-1.0);
struct mint {
	ll x; // typedef long long ll;
	mint(ll x = 0) :x((x%MOD + MOD) % MOD) {}
	mint operator-() const { return mint(-x); }
	mint& operator+=(const mint a) {
		if ((x += a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += MOD - a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator*=(const mint a) {
		(x *= a.x) %= MOD;
		return *this;
	}
	mint operator+(const mint a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint a) const {
		mint res(*this);
		return res *= a;
	}
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}
	// for prime MOD
	mint inv() const {
		return pow(MOD - 2);
	}
	mint& operator/=(const mint a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint a) const {
		mint res(*this);
		return res /= a;
	}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
struct combination {
	vector<mint> fact, ifact;
	combination(int n) :fact(n + 1), ifact(n + 1) {
		assert(n < MOD);
		fact[0] = 1;
		for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
	}
	mint operator()(int n, int k) {
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
};
struct Sieve {
	int n;
	vector<int> f, primes;

	Sieve(int n = 1) :n(n), f(n + 1) {
		f[0] = f[1] = -1;
		for (ll i = 2; i <= n; ++i) {
			if (f[i]) continue;
			primes.push_back(i);
			f[i] = i;
			for (ll j = i * i; j <= n; j += i) {
				if (!f[j]) f[j] = i;
			}
		}
	}

	bool isPrime(int x) { return f[x] == x; }

	vector<int> factorList(int x) {
		vector<int> res;
		while (x != 1) {
			res.push_back(f[x]);
			x /= f[x];
		}
		return res;
	}

	vector<P> factor(int x) {
		vector<int> fl = factorList(x);
		if (fl.size() == 0) return {};
		vector<P> res(1, P(fl[0], 0));
		for (int p : fl) {
			if (res.back().first == p) {
				res.back().second++;
			}
			else {
				res.emplace_back(p, 1);
			}
		}
		return res;
	}
};
template<class t> t gcd(t a, t b) { return b != 0 ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return a / g * b;
}
bool prime(ll n) {
	for (ll i = 2; i <= sqrt(n); i++) {
		if (n%i == 0)return false;
	}
	return n != 1;
}
map<ll, ll> prime_factor(ll n) {
	map<ll, ll> ret;
	for (ll i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			ret[i]++;
			n /= i;
		}
	}
	if (n != 1) ret[n] = 1;
	return ret;
}
ll modinv(ll a, ll m) {
	ll b = m, u = 1, v = 0;
	while (b) {
		ll t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
vector<pair<char, int>> RunLength(string s) {
	if (s.size() == 0)return {};
	vector<pair<char, int>>res(1, pair<char, int>(s[0], 0));
	for (char p : s) {
		if (res.back().first == p) {
			res.back().second++;
		}
		else {
			res.emplace_back(p, 1);
		}
	}
	return res;
}
// Digit Count
int GetDigit(int num) {
	return log10(num) + 1;
}
// bit calculation[how many "1"] (= __builtin_popcount())
int bit_count(int n) {
	int cnt = 0;
	while (n > 0) {
		if (n % 2 == 1)cnt++;
		n /= 2;
	}
	return cnt;
}
vector<long long> enum_divisors(long long N) {
	vector<long long> res;
	for (long long i = 1; i * i <= N; ++i) {
		if (N % i == 0) {
			res.push_back(i);
			if (N / i != i) res.push_back(N / i);
		}
	}
	sort(res.begin(), res.end());
	return res;
}
const ll dx[4] = { 1,0,-1,0 };
const ll dy[4] = { 0,1,0,-1 };
struct edge { ll to, cost; };
typedef long double ld;
using Graph = vector<VE>;
class UnionFind {
public:
	vector <ll> par; // 各元の親を表す配列
	vector <ll> siz; // 素集合のサイズを表す配列(1 で初期化)

	// Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1) {
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.resize(sz_, 1);
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}

	// Member Function
	// Find
	ll root(ll x) { // 根の検索
		while (par[x] != x) {
			x = par[x] = par[par[x]]; // x の親の親を x の親とする
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;
		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		return true;
	}

	bool issame(ll x, ll y) { // 連結判定
		return root(x) == root(y);
	}

	ll size(ll x) { // 素集合のサイズ
		return siz[root(x)];
	}
};



signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	//cout << fixed << setprecision(15);
	//combination com(200010);
	int n; cin >> n;
	vector<P>p;
	rep(i, n) {
		int a, b; cin >> a >> b;
		int tmp = a, tmp2=b;
		a = tmp2 - tmp, b =tmp2 + tmp;
		p.push_back(make_pair(b, a));
	}
	sort(ALL(p));
	int now = 0;
	int ans = 0;
	rep(i, n) {
		if (now <= p[i].second) {
			now = p[i].first; ans++;
		}
	}
	cout << ans << endl;
	return 0;
}