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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <numeric>
#include <vector>
#include <string>
#include <bitset>
#include <queue>
#include <tuple>
#include <regex>
#include <cmath>
#include <map>
#include <cstring>

template<typename T> bool chmax( T &a, const T &b ) { if ( a <= b ) { a = b; return ( true ); } else { return ( false ); } }
template<typename T> bool chmin( T &a, const T &b ) { if ( a >= b ) { a = b; return ( true ); } else { return ( false ); } }

using namespace std;

using ll = long long;
using ull = unsigned long long;

using Pint = pair<int, int>;
using Pll  = pair<ll, ll>;
using Pull = pair<ull, ull>;

#define eb emplace_back
#define pb push_back
#define mp make_pair
#define mt make_tuple

#define F first
#define S second

#define rep( i, n ) for ( int i = 0; i < (int)( n ); ++i )
#define reps( i, n ) for ( int i = 1; i <= (int)( n ); ++i )
#define rrep( i, n ) for ( int i = (int)( ( n ) - 1 ); i >= 0; --i )
#define rreps( i, n ) for ( int i = (int)( ( n ) ); i > 0; --i )
#define arep( i, v ) for ( auto &&i : ( v ) )

template<typename T> T gcd( const T a, const T b ) { return ( b ? gcd( b, a % b ) : a ); }
template<typename T> T lcm( const T a, const T b ) { return ( a / gcd( a, b ) * b ); }

#define ALL( c ) ( c ).begin(), ( c ).end()
#define RALL( c ) ( c ).rbegin(), ( c ).rend()
#define UNIQUE( c ) ( c ).erase( unique( ( c ).begin(), ( c ).end() ), ( c ).end() )

constexpr ll MOD = 1000000007LL;
template<typename T = ll> constexpr T MAX = numeric_limits<T>::max();

#define y0 y3487465
#define y1 y8687969
#define j0 j1347829
#define j1 j234892
#define next asdnext
#define prev asdprev

template<typename T = ll> class UnionFind {
public:
	UnionFind( T n ) { par.resize( n ); siz.resize( n ); rep( i, n ) { par[i] = i; siz[i] = 1; } }
	T find( T x ) { if ( x == par[x] ) return ( x ); else return( par[x] = find( par[x] ) ); }
	void unite( T x, T y ) { T xx = find( x ); T yy = find( y ); if ( xx == yy ) return;
		if ( siz[xx] <= siz[yy] ) swap( xx, yy ); par[yy] = xx; siz[xx] += siz[yy]; }
private:
	vector<T> par, siz;
};

template<typename T = ll> class CompareMax {
public:	T operator()( T a, T b ) { return ( max( a, b ) ); }
};
template<typename T = ll> class CompareMin {
public:	T operator()( T a, T b ) { return ( min( a, b ) ); }
};

template<typename T = ll, typename F = CompareMax<T>, T I = 0> class SegTree {
public:
	SegTree( T n ) { N = n; v.resize( 4 * n ); rep( i, 4 * n ) v[i] = I; }
	void update( T i, T x ) { i += N - 1; v[i] = x; while ( i > 0 ) {
			i = ( i - 1 ) / 2; v[i] = func( v[i * 2 + 1], v[i * 2 + 2] ); } }
	T query( T a, T b, T k, T l, T r ) {
		if ( r <= a || b <= l ) return ( I );
		if ( a <= l && r <= b ) return ( v[k] );
		else { T vl = query( a, b, k * 2 + 1, l, ( l + r ) / 2 );
			T vr = query( a, b, k * 2 + 2, ( l + r ) / 2, r );
			return ( func( vl, vr ) ); }
	}
private:
	T N; F func; vector<T> v;
};

template<typename T = ll> T solveLIS( const vector<T> &v ) {
	vector<T> dp( v.size(), numeric_limits<T>::max() );
	rep( i, v.size() ) {
		*lower_bound( ALL( dp ), v[i] ) = v[i];
	}
	return ( distance( dp.begin(), lower_bound( ALL( dp ), numeric_limits<T>::max() ) ) );
}

template<typename T = ll> T power( T a, T b, T m = MOD ) {
	T res = 1;
	while ( b > 0 ) { if ( b & 1 ) res = res * a % m;
		a = a * a % m; b >>= 1; }
	return ( res );
}

/*
constexpr ll COMBSIZE = 500010;
ll fact[COMBSIZE];
ll inv[COMBSIZE];
ll inv_fact[COMBSIZE];

template<typename T> void initComb( T n, T m = MOD )
{
	fact[0] = fact[1] = inv_fact[0] = inv_fact[1] = 1;
	inv[1] = 1;
	for ( int i = 2; i < n; i++ ) {
		fact[i] = ( fact[i - 1] * i ) % m;
		inv[i] = m - inv[m % i] * ( m / i ) % m;
		inv_fact[i] = inv_fact[i - 1] * inv[i] % m;
	}
}

template<typename T> T comb( T n, T r, T m = MOD )
{
	if ( n < r ) return ( 0 );
	if ( n < 0 || r < 0 ) return ( 0 );
	return ( fact[n] * ( inv_fact[r] * inv_fact[n - r] % m ) % m );
}
*/

/*
	regex reg( R"(^(dream|dreamer|erase|eraser)+$)" );
	smatch m;

	if ( regex_match( s, m, reg ) )
	{
		cout << "YES" << endl;
	}
	else
	{
		cout << "NO" << endl;
	}
*/
/*
	cout << setprecision( 16 );
	cout << fixed << setprecision( 16 );
*/

template<typename T = int> vector<T> Zalgo( const string &S ) {
	T N = (T)S.size();
	vector<T> res(N);
	res[0] = N;
	T i = 1, j = 0;
	while ( i < N ) {
		while (i+j < N && S[j] == S[i+j]) ++j;
		res[i] = j;
		if ( j == 0 ) { ++i; continue; }
		int k = 1;
		while ( i + k < N && k + res[k] < j ) res[i + k] = res[k], ++k;
		i += k, j -= k;
	}
	return ( res );
}

void replace( string &s, string t, string r ) {
	string::size_type p = 0;
	while ( ( p = s.find( t, p ) ) != string::npos ) {
		s.replace( p, t.length(), r );
		p += r.length();
	}
}


int main()
{
	ll N; cin >> N;
	vector<Pll> v( N );
	rep( i, N ) cin >> v[i].F >> v[i].S;
	v.eb( mp( -MOD * 2, 0 ) );
	v.eb( mp( MOD * 2, 0 ) );

	sort( ALL( v ), []( Pll a, Pll b ) {
			return ( a.F < b.F );
		} );

	vector<ll> vv( N + 3 );
	rep( i, N )
	{
		vv[max( v[i].F - v[i].S, 0LL )] += 1;
		vv[min( v[i].F + v[i].S, 100001LL )] -= 1;
	}
	vector<ll> vvv( N + 3 );
	rep( i, N + 2 )
	{
		vvv[i + 1] = vvv[i] + vv[i];
	}
	sort( ALL( vvv ), greater<ll>() );

	cout << N - ( vvv[0] - 1 ) << endl;

	return ( 0 );
}