There is a white sheet of paper lying on a rectangle table. The sheet is a rectangle with its sides parallel to the sides of the table. If you will take a look from above and assume that the bottom left corner of the table has coordinates (0, 0), and coordinate axes are left and bottom sides of the table, then the bottom left corner of the white sheet has coordinates (x_1, y_1), and the top right — (x_2, y_2).

After that two black sheets of paper are placed on the table. Sides of both black sheets are also parallel to the sides of the table. Coordinates of the bottom left corner of the first black sheet are (x_3, y_3), and the top right — (x_4, y_4). Coordinates of the bottom left corner of the second black sheet are (x_5, y_5), and the top right — (x_6, y_6). 

<image> Example of three rectangles.

Determine if some part of the white sheet can be seen from the above after the two black sheets are placed. The part of the white sheet can be seen if there is at least one point lying not strictly inside the white sheet and strictly outside of both black sheets.

Input

The first line of the input contains four integers x_1, y_1, x_2, y_2 (0 ≤ x_1 < x_2 ≤ 10^{6}, 0 ≤ y_1 < y_2 ≤ 10^{6}) — coordinates of the bottom left and the top right corners of the white sheet.

The second line of the input contains four integers x_3, y_3, x_4, y_4 (0 ≤ x_3 < x_4 ≤ 10^{6}, 0 ≤ y_3 < y_4 ≤ 10^{6}) — coordinates of the bottom left and the top right corners of the first black sheet.

The third line of the input contains four integers x_5, y_5, x_6, y_6 (0 ≤ x_5 < x_6 ≤ 10^{6}, 0 ≤ y_5 < y_6 ≤ 10^{6}) — coordinates of the bottom left and the top right corners of the second black sheet.

The sides of each sheet of paper are parallel (perpendicular) to the coordinate axes.

Output

If some part of the white sheet can be seen from the above after the two black sheets are placed, print "YES" (without quotes). Otherwise print "NO".

Examples

Input


2 2 4 4
1 1 3 5
3 1 5 5


Output


NO


Input


3 3 7 5
0 0 4 6
0 0 7 4


Output


YES


Input


5 2 10 5
3 1 7 6
8 1 11 7


Output


YES


Input


0 0 1000000 1000000
0 0 499999 1000000
500000 0 1000000 1000000


Output


YES

Note

In the first example the white sheet is fully covered by black sheets.

In the second example the part of the white sheet can be seen after two black sheets are placed. For example, the point (6.5, 4.5) lies not strictly inside the white sheet and lies strictly outside of both black sheets.
import javafx.util.Pair;

import java.io.*;
import java.math.*;
import java.util.*;
@SuppressWarnings("Duplicates")
// author @mdazmat9
public class Main{
    static Scanner sc = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    public static void main(String[] args) throws IOException {
        int test = 1;
      //  test=sc.nextInt();
        for (int ind = 0; ind < test; ind++) {
            solve();
        }
        out.flush();
    }
    static void solve() {
        Rectangle W = new Rectangle(sc.nextInt(), sc.nextInt(), sc.nextInt(), sc.nextInt());
        Rectangle B1 = new Rectangle(sc.nextInt(), sc.nextInt(), sc.nextInt(), sc.nextInt());
        Rectangle B2 = new Rectangle(sc.nextInt(), sc.nextInt(), sc.nextInt(), sc.nextInt());
        Rectangle X1 = W.intersect(B1);
        Rectangle X2 = W.intersect(B2);
        Rectangle X12 = X1.intersect(X2);
        out.println(X1.area() + X2.area() - X12.area() == W.area() ? "NO" : "YES");
    }
    static class Rectangle {
        int xa, ya, xb, yb;
        public Rectangle(int a, int b, int c, int d) {
            xa = a; ya = b; xb = c; yb = d;
        }

        public Rectangle intersect(Rectangle N) {
            Rectangle intersection = new Rectangle(0, 0, 0, 0);
            // Make the lower left part as upper right as possible
            intersection.xa = Math.max(xa, N.xa);
            intersection.ya = Math.max(ya, N.ya);

            // Make the upper right part as lower left as possible
            intersection.xb = Math.min(xb, N.xb);
            intersection.yb = Math.min(yb, N.yb);
            return intersection;
        }

        public long area() {
            if (xa < xb && ya < yb) {
                long area = (long) (xb - xa) * (yb - ya);
                return area;
            }
            else {
                return 0;
            }
        }
    }




    static int[] intarray(int n){ int [] a=new int[n];for(int i=0;i<n;i++)a[i]=sc.nextInt();return a; }
    static void sort(int[]a){ shuffle(a);Arrays.sort(a);}
    static void sort(long[]a){ shuffle(a);Arrays.sort(a);}
    static long[] longarray(int n){ long [] a=new long[n];for(int i=0;i<n;i++)a[i]=sc.nextLong();return a; }
    static ArrayList<Integer> intlist(int n){ArrayList<Integer> list=new ArrayList<>();for(int i=0;i<n;i++)list.add(sc.nextInt());return list; }
    static ArrayList<Long> longlist(int n){ArrayList<Long> list=new ArrayList<>();for(int i=0;i<n;i++)list.add(sc.nextLong());return list; }
    static int[][] int2darray(int n,int m){ int [][] a=new int[n][m];for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ a[i][j]=sc.nextInt(); } }return a; }
    static long[][] long2darray(int n,int m){ long [][] a=new long[n][m];for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ a[i][j]=sc.nextLong(); } }return a; }
    static char[][] char2darray(int n,int m){ char [][] a=new char[n][m];for(int i=0;i<n;i++){ String s=sc.next(); a[i]=s.toCharArray(); }return a; }

    static double pi=3.14159265358979323846264;
    public static double logb( double a, double b ) {return Math.log(a) / Math.log(b); }
    static long fast_pow(long a, long b,long abs) {
        if(b == 0) return 1L;
        long val = fast_pow(a, b / 2,abs);
        if(b % 2 == 0) return val * val % abs;
        else return val * val % abs * a % abs;
    }
    static long abs = (long)1e9 + 7;
    static   void shuffle(int[] a) { int n = a.length;for(int i = 0; i < n; i++) { int r = i + (int) (Math.random() * (n - i));int tmp = a[i];a[i] = a[r];a[r] = tmp; } }
    static   void shuffle(long[] a) { int n = a.length;for(int i = 0; i < n; i++) { int r = i + (int) (Math.random() * (n - i));long tmp = a[i];a[i] = a[r];a[r] = tmp; } }
    static long gcd(long a , long b) {
        if(b == 0) return a;
        return gcd(b , a % b);
    }
}

class Scanner {
    public BufferedReader reader;
    public StringTokenizer st;
    public Scanner(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
        st = null;
    }
    public String next() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                String line = reader.readLine();
                if (line == null) return null;
                st = new StringTokenizer(line);
            } catch (Exception e) {
                throw (new RuntimeException());
            }
        }
        return st.nextToken();
    }
    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong() {
        return Long.parseLong(next());
    }
    public double nextDouble() {
        return Double.parseDouble(next());
    }
}
