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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        D solver = new D();
        solver.solve(1, in, out);
        out.close();
    }

    static class D {
        int n;
        int[] x;
        int[] y;

        boolean test(long a, long b, long c) {
            Long x1 = null, x2 = null, y1 = null, y2 = null;
            long a2 = 0, b2 = 0, c2 = 0;
            for (int i = 0; i < n; i++) {
                if (a * x[i] + b * y[i] + c != 0) {
                    if (x1 == null) {
                        x1 = (long) x[i];
                        y1 = (long) y[i];
                    } else if (x2 == null) {
                        x2 = (long) x[i];
                        y2 = (long) y[i];
                        a2 = y2 - y1;
                        b2 = x1 - x2;
                        c2 = -a2 * x2 - b2 * y2;
                    } else if (a2 * x[i] + b2 * y[i] + c2 != 0) {
                        return false;
                    }
                }
            }
            return true;
        }

        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            n = in.ni();
            x = new int[n];
            y = new int[n];
            for (int i = 0; i < n; i++) {
                x[i] = in.ni();
                y[i] = in.ni();
            }
            if (n < 5) {
                out.println("YES");
            } else {
                int[] xa = new int[]{x[0], x[0], x[1]};
                int[] xb = new int[]{x[1], x[2], x[2]};
                int[] ya = new int[]{y[0], y[0], y[1]};
                int[] yb = new int[]{y[1], y[2], y[2]};
                for (int i = 0; i < 3; i++) {
                    long a = yb[i] - ya[i];
                    long b = xa[i] - xb[i];
                    long c = -a * xb[i] - b * yb[i];
                    if (test(a, b, c)) {
                        out.println("YES");
                        return;
                    }
                }
                out.println("NO");

            }


        }

    }

    static class FastScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

        public String ns() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    String rl = in.readLine();
                    if (rl == null) {
                        return null;
                    }
                    st = new StringTokenizer(rl);
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int ni() {
            return Integer.parseInt(ns());
        }

    }
}

