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.
from sys import stdin as cin

def main():
    line = cin.readline().strip()
    a = line.split(" ")
    R,C = int(a[0]),int(a[1])
    m = []
    for t in range(0,R):
        line = cin.readline().strip()
        arr = line.split(" ")
        lst = [int(x) for x in arr]
        m.append(lst)
    T = int(input())
    for t in range(0,T):
        n = int(input())
        found  = False
        r,c = 0,C-1
        while(r < R and c >=0 ):
            if(m[r][c] == n):
                print(str(r) +' '+ str(c))
                found = True
                break
            elif m[r][c] < n:
                r = r + 1
            elif m[r][c] > n:
                c = c - 1
        if not found:
            print( str('-1') +' '+str('-1'))
main()