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.
input_sizes = raw_input()
sizes = []
for size in input_sizes.split():
	sizes.append(int(size))
count = 0
mat = []
while count < sizes[0]:
	numbers = []
	text_input = raw_input()
	for num in text_input.split():
		numbers.append(int(num))
	mat.append(numbers)
	count += 1
test_cases  = raw_input()
test_cases = int(test_cases)
count = 0
while count < test_cases : 
	num = raw_input()
	num = int(num)
	row = 0 
	col = sizes[1]-1
	while 1:
		if(row == sizes[0] or  col == -1):
			print "-1 -1"
			break
		if(num == mat[row][col]):
			print str(row)+" "+str(col)
			break
		elif(num > mat[row][col]):
			row += 1
		else:
			col -= 1
	count += 1