You are given n points on Cartesian plane. Every point is a lattice point (i. e. both of its coordinates are integers), and all points are distinct.

You may draw two straight lines (not necessarily distinct). Is it possible to do this in such a way that every point lies on at least one of these lines?

Input

The first line contains one integer n (1 ≤ n ≤ 105) — the number of points you are given.

Then n lines follow, each line containing two integers xi and yi (|xi|, |yi| ≤ 109)— coordinates of i-th point. All n points are distinct.

Output

If it is possible to draw two straight lines in such a way that each of given points belongs to at least one of these lines, print YES. Otherwise, print NO.

Examples

Input

5
0 0
0 1
1 1
1 -1
2 2


Output

YES


Input

5
0 0
1 0
2 1
1 1
2 3


Output

NO

Note

In the first example it is possible to draw two lines, the one containing the points 1, 3 and 5, and another one containing two remaining points.

<image>
import java.util.*;
import java.lang.*;
import java.io.*;

/*
 * 
 * Comments Here
 * 
 */
public class D41 {
	static BufferedReader br;
	static BufferedWriter bw;
	static StringTokenizer st;

	public static void main(String[] args) throws java.lang.Exception {
		/**/
		br = new BufferedReader(new InputStreamReader(System.in));
		bw = new BufferedWriter(new OutputStreamWriter(System.out));
		// Uncomment to read from file
		/*
		 * / File file = new File("2016/A.in"); br = new BufferedReader(new
		 * FileReader(file)); bw = new BufferedWriter(new
		 * OutputStreamWriter(System.out)); /
		 **/

		int n = Integer.parseInt(br.readLine());

		long[] x = new long[n];
		long[] y = new long[n];

		for (int i = 0; i < n; ++i) {
			st = new StringTokenizer(br.readLine());
			x[i] = Long.parseLong(st.nextToken());
			y[i] = Long.parseLong(st.nextToken());
		}

		if (twoLines(x, y)) {
			bw.write("YES\n");
			br.close();
			bw.close();
			return;
		}

		// if we get here, A and B must not be colinear.
		// C must be colinear with either A or B

		long tempx = x[1];
		x[1] = x[2];
		x[2] = tempx;

		long tempy = y[1];
		y[1] = y[2];
		y[2] = tempy;

		// This tries A,C colinear
		if (twoLines(x, y)) {
			bw.write("YES\n");
			br.close();
			bw.close();
			return;
		}

		tempx = x[0];
		x[0] = x[2];
		x[2] = tempx;

		tempy = y[0];
		y[0] = y[2];
		y[2] = tempy;

		// This tries B,C colinear
		if (twoLines(x, y)) {
			bw.write("YES\n");
			br.close();
			bw.close();
			return;
		}
		
		bw.write("NO\n");

		br.close();
		bw.close();
	}

	static boolean isColinear(long[] x, long[] y) {

		if (x.length < 3) {
			return true;
		}

		long rise = y[1] - y[0];
		long run = x[1] - x[0];

		for (int i = 2; i < x.length; ++i) {

			if ((run * (y[i] - y[0])) != (rise * (x[i] - x[0]))) {
				return false;
			}

		}

		return true;

	}

	static boolean twoLines(long[] x, long[] y) {

		int n = x.length;

		if (n <= 4) {
			return true;
		}

		// try A,B colinear
		long rise = y[1] - y[0];
		long run = x[1] - x[0];

		boolean[] AB = new boolean[n];
		AB[0] = true;
		AB[1] = true;

		int not_AB_count = 0;
		for (int i = 2; i < n; ++i) {

			if ((run * (y[i] - y[0])) == (rise * (x[i] - x[0]))) {
				AB[i] = true;
			} else {
				++not_AB_count;
			}

		}

		// now all remaining points should be colinear

		long[] x_notAB = new long[not_AB_count];
		long[] y_notAB = new long[not_AB_count];
		for (int i = 0; i < n; ++i) {
			if (!AB[i]) {
				x_notAB[not_AB_count - 1] = x[i];
				y_notAB[not_AB_count - 1] = y[i];
				--not_AB_count;
			}
		}

		if (isColinear(x_notAB, y_notAB)) {
			return true;
		}

		return false;

	}

}
