Problem: You are given n points with integer coordinates on the plane. Points are given in a way such that there is no triangle, formed by any three of these n points, which area exceeds S.

Alyona tried to construct a triangle with integer coordinates, which contains all n points and which area doesn't exceed 4S, but, by obvious reason, had no success in that. Please help Alyona construct such triangle. Please note that vertices of resulting triangle are not necessarily chosen from n given points.

Input

In the first line of the input two integers n and S (3 ≤ n ≤ 5000, 1 ≤ S ≤ 1018) are given — the number of points given and the upper bound value of any triangle's area, formed by any three of given n points.

The next n lines describes given points: ith of them consists of two integers xi and yi ( - 108 ≤ xi, yi ≤ 108) — coordinates of ith point.

It is guaranteed that there is at least one triple of points not lying on the same line.

Output

Print the coordinates of three points — vertices of a triangle which contains all n points and which area doesn't exceed 4S.

Coordinates of every triangle's vertex should be printed on a separate line, every coordinate pair should be separated by a single space. Coordinates should be an integers not exceeding 109 by absolute value.

It is guaranteed that there is at least one desired triangle. If there is more than one answer, print any of them.

Example

Input

4 1
0 0
1 0
0 1
1 1


Output

-1 0
2 0
0 2

Note

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Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 * @author Don Li
 */
public class AlyonaTriangles {
    void solve() {
        int n = in.nextInt();
        long s = in.nextLong();
        
        long[][] p = new long[n][2];
        for (int i = 0; i < n; i++) {
            p[i][0] = in.nextInt();
            p[i][1] = in.nextInt();
        }
        
        // find the convex hull for the points
        long[][] h = convex_hull(p);
        int m = h.length;
        
        // find the max_area triangle from points lie on the convex hull
        long max = 0;
        int a = -1, b = -1, c = -1;
        int i = 0, j = 1, k = 2;
        while (true) {
            if (i == m) break;
            long cur = area(h[i], h[j], h[k]);
            while (true) {
                while (true) {
                    int nk = (k + 1) % m;
                    long nxt = area(h[i], h[j], h[nk]);
                    if (nxt >= cur) {
                        k = nk;
                        cur = nxt;
                    } else {
                        break;
                    }
                }
                int nj = (j + 1) % m;
                long nxt = area(h[i], h[nj], h[k]);
                if (nxt >= cur) {
                    j = nj;
                    cur = nxt;
                } else {
                    break;
                }
            }
            if (cur > max) {
                max = cur;
                a = i;
                b = j;
                c = k;
            }
            i++;
            if (j == i) j = (j + 1) % m;
            if (k == j) k = (k + 1) % m;
        }
        
        // get the anticomplementary triangle from the max_area triangle
        long[][] res = new long[3][2];
        res[0][0] = -h[a][0] + h[b][0] + h[c][0];
        res[0][1] = -h[a][1] + h[b][1] + h[c][1];
        res[1][0] = h[a][0] - h[b][0] + h[c][0];
        res[1][1] = h[a][1] - h[b][1] + h[c][1];
        res[2][0] = h[a][0] + h[b][0] - h[c][0];
        res[2][1] = h[a][1] + h[b][1] - h[c][1];
        
        for (long[] ps : res) {
            out.printf("%d %d%n", ps[0], ps[1]);
        }
    }
    
    long[][] convex_hull(long[][] p) {
        int n = p.length;
        Arrays.sort(p, (x, y) -> {
            if (x[0] != y[0]) return Long.compare(x[0], y[0]);
            return Long.compare(x[1], y[1]);
        });
        
        int[] ids = new int[n + 1];
        int k = 0;
        for (int i = 0; i < n; i++) {
            while (k > 1 && ccw(p[ids[k - 2]], p[ids[k - 1]], p[i]) <= 0) k--;
            ids[k++] = i;
        }
        
        for (int i = n - 2, t = k; i >= 0; i--) {
            while (k > t && ccw(p[ids[k - 2]], p[ids[k - 1]], p[i]) <= 0) k--;
            ids[k++] = i;
        }
        
        k--;
        
        long[][] res = new long[k][];
        for (int i = 0; i < k; i++) res[i] = p[ids[i]];
        return res;
    }
    
    int ccw(long[] a, long[] b, long[] c) {
        return Long.signum((b[0] - a[0]) * (c[1] - a[1]) - (c[0] - a[0]) * (b[1] - a[1]));
    }
    
    long area(long[] a, long[] b, long[] c) {
        return Math.abs((b[0] - a[0]) * (c[1] - a[1]) - (c[0] - a[0]) * (b[1] - a[1]));
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new AlyonaTriangles().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}