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>

using namespace std;

#define DUMP(x) std::cerr << (#x) << " = " << (x) << "\n"
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define REP(i, k, n) for (int i = (k); i < (int)(n); ++i)
#define ALL(r) r.begin(), r.end()
#define YES puts("YES")
#define Yes puts("Yes")
#define NO puts("NO")
#define No puts("No")
#define IMP puts("IMPOSSIBLE")
#define Imp puts("Impossible")
#define imp puts("impossible")
#define M1 puts("-1")
#define pb push_back
#define eb emplace_back

template <typename T>
ostream &operator<<(ostream &os, vector<T> &v)
{
    os << "{";
    rep(i, v.size()) os << v[i] << (i == (int)v.size() - 1 ? "" : ", ");
    os << "}";
    return os;
}
template <typename T>
T dup(T x, T y)
{
    return (x + y - 1) / y;
};
template <typename A, size_t N, typename T>
inline void arrayFill(A (&array)[N], const T &val) { std::fill((T *)array, (T *)(array + N), val); }
struct in
{
    const size_t n = 0;
    in() = default;
    in(size_t n) : n(n){};
    template <typename T>
    operator T()
    {
        T ret;
        cin >> ret;
        return ret;
    }
    template <typename T>
    operator vector<T>()
    {
        assert(n != 0);
        vector<T> ret(n);
        for (auto &x : ret)
            cin >> x;
        return ret;
    }
};

using ll = int64_t;
using vint = vector<int32_t>;
using vvint = vector<vint>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vstr = vector<string>;
using pint = pair<int32_t, int32_t>;
using vpint = vector<pint>;
using pll = pair<ll, ll>;
using setint = set<int32_t>;
using setstr = set<string>;
using qint = queue<int32_t>;
using qpint = queue<pint>;

constexpr std::int32_t INF = 1001001001;
constexpr std::int64_t LINF = 1001001001001001001;

void Main()
{
    ll n = in();
    vector<pll> v(n);
    rep(i, n)
    {
        ll x = in(), l = in();
        v[i] = make_pair(x - l, x + l);
    }
    sort(ALL(v), [](auto lhs, auto rhs) {
        return lhs.second == rhs.second ? lhs.first < rhs.first : lhs.second < rhs.second;
    });

    ll cnt = 0;
    ll prev_right = -LINF;
    rep(i, n)
    {
        if (prev_right <= v[i].first)
        {
            ++cnt;
            prev_right = v[i].second;
        }
    }
    cout << cnt << endl;
}

signed main()
{
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << std::fixed << std::setprecision(15);
    Main();
    return 0;
}