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
def h(x, p):
    return x % p


class CodeforcesTask447ASolution:
    def __init__(self):
        self.result = ''
        self.p_n = []
        self.inserts = []

    def read_input(self):
        self.p_n = [int(x) for x in input().split(" ")]
        self.inserts = [int(input()) for x in range(self.p_n[1])]

    def process_task(self):
        hashes = [False for x in range(self.p_n[0])]
        result = -1
        for i, x in enumerate(self.inserts):
            if hashes[h(x, self.p_n[0])]:
                if result < 0:
                    result = i + 1
            else:
                hashes[h(x, self.p_n[0])] = True
        self.result = str(result)

    def get_result(self):
        return self.result


if __name__ == "__main__":
    Solution = CodeforcesTask447ASolution()
    Solution.read_input()
    Solution.process_task()
    print(Solution.get_result())
