Sherlock Holmes loves mind palaces! We all know that.

A mind palace, according to Mr. Holmes is something that lets him retrieve a given memory in the least time posible. For this, he structures his mind palace in a very special way. Let a NxM Matrix denote the mind palace of Mr. Holmes. For fast retrieval he keeps each row and each column sorted. Now given a memory X, you have to tell the position of the memory in Sherlock's mind palace.

Input
Input begins with a line containing space separated N and M.
The next N lines each contain M numbers, each referring to a memory Y.
The next line contains Q, the number of queries.
The next Q lines contain a single element X, the memory you have to search in Sherlock's mind palace.

Output
If Y is present in Mr. Holmes memory, output its position (0-based indexing).
Else output "-1 -1" (quotes for clarity only).

Constraints
2 ≤ N,M ≤ 1000
2 ≤ Q ≤ 1000
-10^9 ≤ X,Y ≤ 10^9

Note : Large Input Files. Use faster I/O methods.

SAMPLE INPUT
5 5
-10 -5 -3 4 9
-6 -2 0 5 10
-4 -1 1 6 12
2 3 7 8 13
100 120 130 140 150
3
0
-2
170

SAMPLE OUTPUT
1 2
1 1
-1 -1

Explanation

The sample is self-explanatory.
import sys
import time
def rowBinSearch(A,row,low,high,key):
	#for i in xrange(low,high+1):
	#	print A[row][i]
	while(low<=high):
		mid=(low+high)/2
		if(A[row][mid]==key):
			return (True,row,mid)
		elif(A[row][mid]>key):
			high=mid-1
		else:
			low=mid+1
	return False,0,0

def colBinSearch(A,col,low,high,key):
	#for i in xrange(low,high+1):
	#	print A[i][col]
	while(low<=high):
		mid=(low+high)/2
		if(A[mid][col]==key):
			return (True,mid,col)
		elif(A[mid][col]>key):
			high=mid-1
		else:
			low=mid+1
	return False,0,0


temp=map(int,raw_input().split())
N=temp[0]
M=temp[1]
A=[]

for _ in xrange(N):
	A.append(map(int,raw_input().split()))
start=time.time()
for _ in xrange(input()):
	found=False
	key=input()
	row=0
	col=M-1
	while(col>=0 and row<=N-1):
		if A[row][col]==key:
			found=True
			print row,col
			break
		elif A[row][col]>key:
			found,rowPos,colPos=rowBinSearch(A,row,0,col-1,key)
			if found:
				print rowPos,colPos
				break

		elif A[row][col]<key:
			found,rowPos,colPos=colBinSearch(A,col,row+1,N-1,key)
			if found:
				print rowPos,colPos
				break

		row=row+1
		col=col-1

	if found!=True:
		print "-1 -1"
end=time.time()
#print "diagonal bineary search"+str(end-start)