DZY has a hash table with p buckets, numbered from 0 to p - 1. He wants to insert n numbers, in the order they are given, into the hash table. For the i-th number xi, DZY will put it into the bucket numbered h(xi), where h(x) is the hash function. In this problem we will assume, that h(x) = x mod p. Operation a mod b denotes taking a remainder after division a by b.

However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the i-th insertion, you should output i. If no conflict happens, just output -1.

Input

The first line contains two integers, p and n (2 ≤ p, n ≤ 300). Then n lines follow. The i-th of them contains an integer xi (0 ≤ xi ≤ 109).

Output

Output a single integer — the answer to the problem.

Examples

Input

10 5
0
21
53
41
53


Output

4


Input

5 5
0
1
2
3
4


Output

-1
from sys import stdin, stdout

def trace(f):
    def traced(*args, **kw):
        print("calling {} with args {}, {}".format(f.__name__, args, kw))
        return f(*args, **kw)
    return traced

@trace
def f(x):
    print(x)

def read():
    return stdin.readline()[:-1]

def readarr(sep=None, maxsplit=-1):
    return read().split(sep, maxsplit)

def readint():
    return int(read())

def readintarr(sep=None, maxsplit=-1):
    return [int(a) for a in readarr(sep, maxsplit)]

def write(*args, **kwargs):
    sep = kwargs.get('sep', ' ')
    end = kwargs.get('end', '\n')
    stdout.write(sep.join(str(a) for a in args) + end)

def writearr(arr, sep=' ', end='\n'):
    stdout.write(sep.join(str(a) for a in arr) + end)

p, n = readintarr()
s = set()
for i in range(1, n+1):
    x = readint() % p
    if x in s:
        write(i)
        break
    s.add(x)
else:
    write(-1)
