You are given a rectangle grid. That grid's size is n × m. Let's denote the coordinate system on the grid. So, each point on the grid will have coordinates — a pair of integers (x, y) (0 ≤ x ≤ n, 0 ≤ y ≤ m).

Your task is to find a maximum sub-rectangle on the grid (x1, y1, x2, y2) so that it contains the given point (x, y), and its length-width ratio is exactly (a, b). In other words the following conditions must hold: 0 ≤ x1 ≤ x ≤ x2 ≤ n, 0 ≤ y1 ≤ y ≤ y2 ≤ m, <image>.

The sides of this sub-rectangle should be parallel to the axes. And values x1, y1, x2, y2 should be integers.

<image>

If there are multiple solutions, find the rectangle which is closest to (x, y). Here "closest" means the Euclid distance between (x, y) and the center of the rectangle is as small as possible. If there are still multiple solutions, find the lexicographically minimum one. Here "lexicographically minimum" means that we should consider the sub-rectangle as sequence of integers (x1, y1, x2, y2), so we can choose the lexicographically minimum one.

Input

The first line contains six integers n, m, x, y, a, b (1 ≤ n, m ≤ 109, 0 ≤ x ≤ n, 0 ≤ y ≤ m, 1 ≤ a ≤ n, 1 ≤ b ≤ m).

Output

Print four integers x1, y1, x2, y2, which represent the founded sub-rectangle whose left-bottom point is (x1, y1) and right-up point is (x2, y2).

Examples

Input

9 9 5 5 2 1


Output

1 3 9 7


Input

100 100 52 50 46 56


Output

17 8 86 92
//Created by Aminul on 4/10/2020.

import java.io.*;
import java.util.*;

import static java.lang.Math.*;

public class I_2 {
    static int n, m, x, y, a, b;

    public static void main(String[] args) throws Exception {
        Scanner in = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        n = in.nextInt();
        m = in.nextInt();
        x = in.nextInt();
        y = in.nextInt();
        a = in.nextInt();
        b = in.nextInt();
        int g = gcd(a, b);
        a /= g;
        b /= g;

        long res[] = solve(0, max(n, m));
        pw.println(res[0] + " " + res[1] + " " + res[2] + " " + res[3]);

        pw.close();
    }

    static long[] brute(long l, long r) {
        while (r >= l) {
           // debug(r, check(r));
            if (check(r)) {
                long[] res = new long[]{lastValidX1, lastValidY1, lastValidX1 + r * a, lastValidY1 + r * b};
                return res;
            }
            r--;
        }
        return null;
    }

    static long[] solve(long l, long r) {
        long mid, res[] = null, temp[];
        while (l <= r) {
            mid = (l + r) >> 1;
            if (check(mid)) {
                res = new long[]{lastValidX1, lastValidY1, lastValidX1 + mid * a, lastValidY1 + mid * b};
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return res;
    }

    static long lastValidX1, lastValidY1;

    static boolean check(long factor) {
        lastValidX1 = findX1(max(0, x - factor * a), min(n, x + factor * a), factor);
        lastValidY1 = findY1(max(0, y - factor * b), min(m, y + factor * b), factor);
        return lastValidX1 != -1 && lastValidY1 != -1;
    }

    static final long INF = (long) 1e18;

    static long findX1(long l, long r, long factor) { //use long/int/double where necessary
        long m1, m2;
        while (r - l > 5) {
            m1 = (2 * l + r) / 3;
            m2 = (l + 2 * r) / 3;
            long p = INF, q = INF;
            {
                long x1 = m1, x2 = x1 + factor * a;
                if (x1 >= 0 && x2 <= n && x1 <= x && x <= x2) {
                    p = max(abs(x1 - x), abs(x2 - x));
                }
            }
            {
                long x1 = m2, x2 = x1 + factor * a;
                if (x1 >= 0 && x2 <= n && x1 <= x && x <= x2) {
                    q = max(abs(x1 - x), abs(x2 - x));
                }
            }



            if (p > q) l = m1;
            else r = m2;
        }
        long min = INF, res = -1;
        for (; l <= r; l++) {
            long x1 = l, x2 = x1 + factor * a, p = INF;
            if (x1 >= 0 && x2 <= n && x1 <= x && x <= x2) {
                p = max(abs(x1 - x), abs(x2 - x));
            }
            if (p < min) {
                min = p;
                res = x1;
            }
        }
        return res;
    }


    static long findY1(long l, long r, long factor) { //use long/int/double where necessary
        long m1, m2;
        while (r - l > 5) {
            m1 = (2 * l + r) / 3;
            m2 = (l + 2 * r) / 3;
            long p = INF, q = INF;
            {
                long y1 = m1, y2 = y1 + factor * b;
                if (y1 >= 0 && y2 <= m && y1 <= y && y <= y2) {
                    p = max(abs(y1 - y), abs(y2 - y));
                }
            }
            {
                long y1 = m2, y2 = y1 + factor * b;
                if (y1 >= 0 && y2 <= m && y1 <= y && y <= y2) {
                    q = max(abs(y1 - y), abs(y2 - y));
                }
            }
            if (p > q) l = m1;
            else r = m2;
        }
        long min = INF, res = -1;
        for (; l <= r; l++) {
            long y1 = l, y2 = y1 + factor * b, p = INF;
            if (y1 >= 0 && y2 <= m && y1 <= y && y <= y2) {
                p = max(abs(y1 - y), abs(y2 - y));
            }
            if (p < min) {
                min = p;
                res = y1;
            }
        }
        return res;
    }


    static int boolToInt(boolean b) {
        return b ? 1 : 0;
    }

    static boolean isValid(long x1, long x2, long y1, long y2) {
        return x1 >= 0 && x2 >= 0 && x1 <= x2 && x2 <= n && y1 >= 0 && y2 >= 0 && y1 <= y2 && y2 <= m
                && x1 <= x && x <= x2 && y1 <= y && y <= y2;
    }

    static int gcd(int p, int q) {
        if (q == 0) return p;
        else return gcd(q, p % q);
    }


    static void debug(Object... obj) {
        System.err.println(Arrays.deepToString(obj));
    }
}
          	  		 		 				 	    	