The clique problem is one of the most well-known NP-complete problems. Under some simplification it can be formulated as follows. Consider an undirected graph G. It is required to find a subset of vertices C of the maximum size such that any two of them are connected by an edge in graph G. Sounds simple, doesn't it? Nobody yet knows an algorithm that finds a solution to this problem in polynomial time of the size of the graph. However, as with many other NP-complete problems, the clique problem is easier if you consider a specific type of a graph.

Consider n distinct points on a line. Let the i-th point have the coordinate xi and weight wi. Let's form graph G, whose vertices are these points and edges connect exactly the pairs of points (i, j), such that the distance between them is not less than the sum of their weights, or more formally: |xi - xj| ≥ wi + wj.

Find the size of the maximum clique in such graph.

Input

The first line contains the integer n (1 ≤ n ≤ 200 000) — the number of points.

Each of the next n lines contains two numbers xi, wi (0 ≤ xi ≤ 109, 1 ≤ wi ≤ 109) — the coordinate and the weight of a point. All xi are different.

Output

Print a single number — the number of vertexes in the maximum clique of the given graph.

Examples

Input

4
2 3
3 1
6 1
0 2


Output

3

Note

If you happen to know how to solve this problem without using the specific properties of the graph formulated in the problem statement, then you are able to get a prize of one million dollars!

The picture for the sample test.

<image>
import static java.lang.Long.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;

import java.util.*;

public class B {

	public B () {
		int N = sc.nextInt();
		long [][] P = sc.nextLongs(N);

		sort(P, new Comparator<long[]>() {
			public int compare(long[] p, long[] q) {
				return signum(p[0] - q[0]);
			}});

		TreeMap<Long, Integer> K = new TreeMap<>();
		for (int i : rep(N)) {
			Long Q = K.floorKey(P[i][0] - P[i][1]);
			long x = P[i][0] + P[i][1];
			int y = Q != null ? K.get(Q) + 1 : 1;

			Long A = K.floorKey(x);
			if (A != null && K.get(A) >= y)
				continue;

			for (;;) {
				Long B = K.ceilingKey(x);
				if (B != null && K.get(B) <= y)
					K.remove(B);
				else
					break;
			}

			K.put(x, y);
		}

		int res = 0;
		for (int y : K.values())
			res = max(res, y);

		exit(res);
	}

	private static int [] rep(int N) { return rep(0, N); }
	private static int [] rep(int S, int T) { if (T <= S) return new int [0]; int [] res = new int [T-S]; for (int i = S; i < T; ++i) res[i-S] = i; return res; }
	////////////////////////////////////////////////////////////////////////////////////
	private final static IOUtils.MyScanner sc = new IOUtils.MyScanner();
	private static void exit (Object o, Object ... A) { IOUtils.print(o, A); IOUtils.exit(); }
	private static class IOUtils {
		public static class MyScanner {
			public String next() { newLine(); return line[index++]; }
			public int nextInt() { return Integer.parseInt(next()); }
			public String nextLine() { line = null; return readLine(); }
			public String [] nextStrings() { return split(nextLine()); }
			public long [] nextLongs() {
				String [] L = nextStrings();
				long [] res = new long [L.length];
				for (int i = 0; i < L.length; ++i)
					res[i] = Long.parseLong(L[i]);
				return res;
			}
			public long [][] nextLongs (int N) {
				long [][] res = new long [N][];
				for (int i = 0; i < N; ++i)
					res[i] = nextLongs();
				return res;
			}
			//////////////////////////////////////////////
			private boolean eol() { return index == line.length; }
			private String readLine() {
				try {
					return r.readLine();
				} catch (Exception e) {
					throw new Error (e);
				}
			}
			private final java.io.BufferedReader r;
			private MyScanner () { this(new java.io.BufferedReader(new java.io.InputStreamReader(System.in))); }
			private MyScanner (java.io.BufferedReader r) {
				try {
					this.r = r;
					while (!r.ready())
						Thread.sleep(1);
					start();
				} catch (Exception e) {
					throw new Error(e);
				}
			}
			private String [] line;
			private int index;
			private void newLine() {
				if (line == null || eol()) {
					line = split(readLine());
					index = 0;
				}
			}
			private String [] split(String s) { return s.length() > 0 ? s.split(" ") : new String [0]; }
		}
		private static String build(Object o, Object ... A) { return buildDelim(" ", o, A); }
		private static String buildDelim(String delim, Object o, Object ... A) {
			StringBuilder b = new StringBuilder();
			append(b, o, delim);
			for (Object p : A)
				append(b, p, delim);
			return b.substring(delim.length());
		}
		//////////////////////////////////////////////////////////////////////////////////
		private static void start() { if (t == 0) t = millis(); }
		private static void append(StringBuilder b, Object o, String delim) {
			if (o.getClass().isArray()) {
				int len = java.lang.reflect.Array.getLength(o);
				for (int i = 0; i < len; ++i)
					append(b, java.lang.reflect.Array.get(o, i), delim);
			} else if (o instanceof Iterable<?>)
				for (Object p : (Iterable<?>) o)
					append(b, p, delim);
			else {
				if (o instanceof Double)
					o = new java.text.DecimalFormat("#.############").format(o);
				b.append(delim).append(o);
			}
		}
		private static java.io.PrintWriter pw = new java.io.PrintWriter(System.out);
		private static void print(Object o, Object ... A) { pw.println(build(o, A)); }
		private static void err(Object o, Object ... A) { System.err.println(build(o, A)); }
		private static void exit() {
			IOUtils.pw.close();
			System.out.flush();
			err("------------------");
			err(IOUtils.time());
			System.exit(0);
		}
		private static long t;
		private static long millis() { return System.currentTimeMillis(); }
		private static String time() { return "Time: " + (millis() - t) / 1000.0; }
	}
	public static void main (String[] args) { new B(); IOUtils.exit(); }
}
