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.
n,m = map(int,raw_input().split())
dime = []

def saddleback_search(dime,y):
	row,col = 0,m-1
	while 0<= row < n and 0<= col < m:
		if dime[row][col] < y:
			row += 1
		elif dime[row][col] > y:
			col -= 1
		elif dime[row][col] == y:
			return row,col
	return -1,-1

for i in xrange(n):
	row = map(int,raw_input().split())
	dime.append(row)
q = input()
for i in xrange(q):
	y = input()
	t1,t2 = saddleback_search(dime,y)
	print "%d %d" % (t1,t2)