Consider the following game for two players. There is one white token and some number of black tokens. Each token is placed on a plane in a point with integer coordinates x and y.

The players take turn making moves, white starts. On each turn, a player moves all tokens of their color by 1 to up, down, left or right. Black player can choose directions for each token independently.

After a turn of the white player the white token can not be in a point where a black token is located. There are no other constraints on locations of the tokens: positions of black tokens can coincide, after a turn of the black player and initially the white token can be in the same point with some black point. If at some moment the white player can't make a move, he loses. If the white player makes 10100500 moves, he wins.

You are to solve the following problem. You are given initial positions of all black tokens. It is guaranteed that initially all these positions are distinct. In how many places can the white token be located initially so that if both players play optimally, the black player wins?

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of black points.

The (i + 1)-th line contains two integers xi, yi ( - 105 ≤ xi, yi, ≤ 105) — the coordinates of the point where the i-th black token is initially located.

It is guaranteed that initial positions of black tokens are distinct.

Output

Print the number of points where the white token can be located initially, such that if both players play optimally, the black player wins.

Examples

Input

4
-2 -1
0 1
0 -3
2 -1


Output

4


Input

4
-2 0
-1 1
0 -2
1 -1


Output

2


Input

16
2 1
1 2
-1 1
0 1
0 0
1 1
2 -1
2 0
1 0
-1 -1
1 -1
2 2
0 -1
-1 0
0 2
-1 2


Output

4

Note

In the first and second examples initial positions of black tokens are shown with black points, possible positions of the white token (such that the black player wins) are shown with white points.

The first example: <image>

The second example: <image>

In the third example the white tokens should be located in the inner square 2 × 2, to make the black player win. <image>
import java.io.*;
import java.util.*;

public class MainD {
	static final StdIn in = new StdIn();
	static final PrintWriter out = new PrintWriter(System.out);
	static final int e5=(int)1e5;
	
	public static void main(String[] args) {
		int n=in.nextInt();
		List<Pair> pts1 = new ArrayList<Pair>(), pts2 = new ArrayList<Pair>();
		for(int i=0; i<n; ++i) {
			int x=in.nextInt(), y=in.nextInt();
			if((x+y)%2==0)
				pts1.add(new Pair((x+y)/2+e5, (x-y)/2+e5));
			else
				pts2.add(new Pair((x+1+y)/2+e5, (x+1-y)/2+e5));
		}
		out.println(solve(pts1)+solve(pts2));
		out.close();
	}
	
	static long solve(List<Pair> pts) {
		int n=pts.size();
		int[] pmn1 = new int[2*e5+1], pmx1 = new int[2*e5+1], pmn2 = new int[2*e5+1], pmx2 = new int[2*e5+1];
		Arrays.fill(pmn1, 2*e5);
		Arrays.fill(pmn2, 2*e5);
		for(Pair p : pts) {
			pmn1[p.a]=Math.min(p.b, pmn1[p.a]);
			pmx1[p.a]=Math.max(p.b, pmx1[p.a]);
			pmn2[p.a]=Math.min(p.b, pmn2[p.a]);
			pmx2[p.a]=Math.max(p.b, pmx2[p.a]);
		}
		long t=0;
		for(int i=1; i<=2*e5; ++i) {
			pmn1[i]=Math.min(pmn1[i-1], pmn1[i]);
			pmx1[i]=Math.max(pmx1[i-1], pmx1[i]);
		}
		for(int i=2*e5; i>0; --i) {
			t+=Math.max(Math.min(pmx1[i-1], pmx2[i])-Math.max(pmn1[i-1], pmn2[i]), 0);
			pmn2[i-1]=Math.min(pmn2[i], pmn2[i-1]);
			pmx2[i-1]=Math.max(pmx2[i], pmx2[i-1]);
		}
		return t;
	}
	
	static class Pair {
		int a, b;
		Pair(int a, int b) {
			this.a=a;
			this.b=b;
		}
	}
	
	static class StdIn {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;
		public StdIn() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}
		public StdIn(InputStream in) {
			try{
				din = new DataInputStream(in);
			} catch(Exception e) {
				throw new RuntimeException();
			}
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}
		public String next() {
			int c;
			while((c=read())!=-1&&(c==' '||c=='\n'||c=='\r'));
			StringBuilder s = new StringBuilder();
			while (c != -1)
			{
				if (c == ' ' || c == '\n'||c=='\r')
					break;
				s.append((char)c);
				c=read();
			}
			return s.toString();
		}
		public String nextLine() {
			int c;
			while((c=read())!=-1&&(c==' '||c=='\n'||c=='\r'));
			StringBuilder s = new StringBuilder();
			while (c != -1)
			{
				if (c == '\n'||c=='\r')
					break;
				s.append((char)c);
				c = read();
			}
			return s.toString();
		}
		public int nextInt() {
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do
				ret = ret * 10 + c - '0';
			while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}
		public int[] readIntArray(int n, int os) {
			int[] ar = new int[n];
			for(int i=0; i<n; ++i)
				ar[i]=nextInt()+os;
			return ar;
		}
		public long nextLong() {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do
				ret = ret * 10 + c - '0';
			while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}
		public long[] readLongArray(int n, long os) {
			long[] ar = new long[n];
			for(int i=0; i<n; ++i)
				ar[i]=nextLong()+os;
			return ar;
		}
		public double nextDouble() {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do
				ret = ret * 10 + c - '0';
			while ((c = read()) >= '0' && c <= '9');
			if (c == '.')
				while ((c = read()) >= '0' && c <= '9')
					ret += (c - '0') / (div *= 10);
			if (neg)
				return -ret;
			return ret;
		}
		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}
		private byte read() {
			try{
				if (bufferPointer == bytesRead)
					fillBuffer();
				return buffer[bufferPointer++];
			} catch(IOException e) {
				throw new RuntimeException();
			}
		}
		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}
	}
}