# Time:  O(k)
# Space: O(1)

class Solution(object):
    def smallestRepunitDivByK(self, K):
        """
        :type K: int
        :rtype: int
        """
        # by observation, K % 2 = 0 or K % 5 = 0, it is impossible
        if K % 2 == 0 or K % 5 == 0:
            return -1

        # let f(N) is a N-length integer only containing digit 1
        # if there is no N in range (1..K) s.t. f(N) % K = 0
        # => there must be K remainders of f(N) % K in range (1..K-1) excluding 0
        # => due to pigeonhole principle, there must be at least 2 same remainders
        # => there must be some x, y in range (1..K) and x > y s.t. f(x) % K = f(y) % K
        # => (f(x) - f(y)) % K = 0
        # => (f(x-y) * 10^y) % K = 0
        # => due to (x-y) in range (1..K)
        # => f(x-y) % K != 0
        # => 10^y % K = 0
        # => K % 2 = 0 or K % 5 = 0
        # => -><-
        # it proves that there must be some N in range (1..K) s.t. f(N) % K = 0
        result = 0
        for N in xrange(1, K+1):
            result = (result*10+1) % K
            if not result:
                return N
        assert(False)
        return -1  # never reach