You've got two rectangular tables with sizes na × ma and nb × mb cells. The tables consist of zeroes and ones. We will consider the rows and columns of both tables indexed starting from 1. Then we will define the element of the first table, located at the intersection of the i-th row and the j-th column, as ai, j; we will define the element of the second table, located at the intersection of the i-th row and the j-th column, as bi, j. 

We will call the pair of integers (x, y) a shift of the second table relative to the first one. We'll call the overlap factor of the shift (x, y) value:

<image>

where the variables i, j take only such values, in which the expression ai, j·bi + x, j + y makes sense. More formally, inequalities 1 ≤ i ≤ na, 1 ≤ j ≤ ma, 1 ≤ i + x ≤ nb, 1 ≤ j + y ≤ mb must hold. If there are no values of variables i, j, that satisfy the given inequalities, the value of the sum is considered equal to 0. 

Your task is to find the shift with the maximum overlap factor among all possible shifts.

Input

The first line contains two space-separated integers na, ma (1 ≤ na, ma ≤ 50) — the number of rows and columns in the first table. Then na lines contain ma characters each — the elements of the first table. Each character is either a "0", or a "1".

The next line contains two space-separated integers nb, mb (1 ≤ nb, mb ≤ 50) — the number of rows and columns in the second table. Then follow the elements of the second table in the format, similar to the first table.

It is guaranteed that the first table has at least one number "1". It is guaranteed that the second table has at least one number "1".

Output

Print two space-separated integers x, y (|x|, |y| ≤ 109) — a shift with maximum overlap factor. If there are multiple solutions, print any of them.

Examples

Input

3 2
01
10
00
2 3
001
111


Output

0 1


Input

3 3
000
010
000
1 1
1


Output

-1 -1


import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Problem62 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		out=new PrintWriter (new BufferedOutputStream(System.out));
		FastReader s=new FastReader();
		int na=s.nextInt();
		int ma=s.nextInt();
		int[][] a=new int[na][ma];
		for(int i=0;i<na;i++) {
			String temp=s.next();
			for(int j=0;j<temp.length();j++) {
				a[i][j]=temp.charAt(j)-'0';
			}
		}
		int nb=s.nextInt();
		int mb=s.nextInt();
		int[][] b=new int[nb][mb];
		for(int i=0;i<nb;i++) {
			String temp=s.next();
			for(int j=0;j<temp.length();j++) {
				b[i][j]=temp.charAt(j)-'0';
			}
		}
		long ans=0;
		int x=0,y=0;
		int first=Integer.max(na,nb);
		int second=Integer.max(ma, mb);
		for(int i=-first;i<first;i++) {
			for(int j=-second;j<second;j++) {
				int temp=multiply(a,b,i,j);
				if(temp>ans) {
					x=i;
					y=j;
					ans=temp;
				}
			}
		}
		out.println(x+" "+y);
		out.close();
	}

	public static int multiply(int[][] a,int[][] b,int x,int y) {
		int ans=0;
		for(int i=0;i<a.length;i++) {
			for(int j=0;j<a[i].length;j++) {
				if(i+x<b.length && j+y<b[0].length && i+x>=0 && j+y>=0) {
					ans+=a[i][j]*b[i+x][j+y];
				}
			}
		}
		return ans;
	}
	
	public static PrintWriter out;
	
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception e) {
				e.printStackTrace();
			}
			return str;
		}
	}
	
}
