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: Accepted)

#include <bits/stdc++.h>
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
using namespace std;

// #define int long long
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

ll mod = (1000000007LL);

inline ll poww(ll a, ll b)
{
    ll res = 1;
    while (b > 0)
    {
        if(b & 1) res = (res * a)%mod;
        a = (a * a)%mod;
        b >>= 1;
    }
    return res;
}
inline pii operator+(pii A, pii B){
	return pii(A.f + B.f, A.s + B.s);
}
inline pii operator-(pii A, pii B){
	return pii(A.f - B.f, A.s - B.s);
}
inline ll mult(ll a, ll b){
	return (a * b) % mod;
}

ll gcd (ll a, ll b) {
    while (b) {
        a %= b;
        swap(a, b);
    }
    return a;
}

/*
	>Colocar (>DEFINE INT LONG LONG<) no inicio do template
	>mod = (1e9 + 7), por padrao (NAO ESQUECER DE ALTERAR)
	>NAO ESQUECER DEFINE INT LONG LONG
	> N = 1 ? 
*/

///CUIDADO COM MAXN
#define N 300050

int n, m, q, k, v[N];

pii TT, w[N];

string s;
vector<pii> line;
int dp[N];
int solve(int i){
    if(i >= n) return 0;
    if(dp[i] != -1) return dp[i];
    int ini = i, fim = n - 1, mid, best=n;
    while(fim >= ini){
        mid = (ini + fim)/2;
        if(line[mid].f >= line[i].s)
            best = mid, fim = mid - 1;
        else ini = mid + 1;
    }
    return dp[i] =max(solve(i + 1), solve(best) + 1);
}
int32_t main()
{
	ios::sync_with_stdio(false); cin.tie(0);
    memset(dp, -1,sizeof dp);
    cin>>n;
    for(int i = 1; i <= n; i++){
        int x, l;
        cin>>x>>l;
        line.pb({x - l, x + l});
    }
    sort(all(line));

    cout<<solve(0)<<"\n";
}