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.
a = list(map(lambda x: int(x), raw_input().split()))
a0 = a[0]
a1 = a[1]
input_list = [[] for i in range(a0)]
for i in range(a0):
	input_list[i] = list(map(lambda x: int(x), raw_input().split()))

b = input()
list_num = []
for i in range(b):
	c = input()
	list_num.append(c)


def locate(obj,lst):
	x = 0
	y = a0 - 1
	while y >= 0  and x <= a1 - 1:
		tem = lst[y][x]
		if obj == tem:
			return y,x
		elif obj < tem:
			y -= 1
		else:
			x += 1
	return (-1, -1)
	
for i in list_num:
	y,x = locate(i,input_list)
	print str(y) + " " + str(x)