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.
# Read input from stdin and provide input before running code

dim = raw_input().split()
m,n = int(dim[0]),int(dim[1])
arr = []
queries = []
for i in range(m):
	row = raw_input()
	arr.append(map(int, row.split()))
query_count = int(raw_input())

for i in range(query_count):
	queries.append(int(raw_input()))
	
def checkElement(item, row, col, m, n):
	r = row
	c = col
	while True:
		if item == arr[r][c]:
			return r,c
		elif item < arr[r][c]:
			c = c-1
		elif item > arr[r][c]:
			r = r+1
		if c < 0 or r > m:
			return -1,-1
			
for q in queries:
	r,c = checkElement(q, 0, n-1, m-1, n-1)
	print('%d %d' % (r, c))