human
Andrew, Fedor and Alex are inventive guys. Now they invent the game with strings for two players.

Given a group of n non-empty strings. During the game two players build the word together, initially the word is empty. The players move in turns. On his step player must add a single letter in the end of the word, the resulting word must be prefix of at least one string from the group. A player loses if he cannot move.

Andrew and Alex decided to play this game k times. The player who is the loser of the i-th game makes the first move in the (i + 1)-th game. Guys decided that the winner of all games is the player who wins the last (k-th) game. Andrew and Alex already started the game. Fedor wants to know who wins the game if both players will play optimally. Help him.

Input

The first line contains two integers, n and k (1 ≤ n ≤ 105.
1 ≤ k ≤ 109).

Each of the next n lines contains a single non-empty string from the given group. The total length of all strings from the group doesn't exceed 105. Each string of the group consists only of lowercase English letters.

Output

If the player who moves first wins, print "First", otherwise print "Second" (without the quotes).

Examples

Input

2 3
a
b

Output

First

Input

3 1
a
b
c

Output

First

Input

1 2
ab

Output

Second
The above is tricky. Write me a correct solution in PYTHON.

=====
Can you provide a Python program to fix of at least one string from the group?
assistant
#Here is Python program to fix of at least one string from the group.

'''input
1 2
ab
'''

n,k = map(int, raw_input().split())
words = [raw_input().strip() for i in range(n)]

class Node():

    def __init__(self, value=''):
        self.val = value
        self.end = 0            # Does a word end at that letter
        self.child = {}

    def add(self, new):           # new is a new node
        self.child[new] = Node(new)

# Let's make a trie baby
head = Node()

for word in words:
    curr = head
    for pos, letter in enumerate(word):
        if letter not in curr.child:
            curr.add(letter)
        curr = curr.child[letter]
        if pos == len(word)-1:
            curr.end = 1

from collections import defaultdict
curr = head

win, lose = defaultdict(bool), defaultdict(bool)
def prinwords(curr, head):
    if head.end==1:
        win[curr] = False
        lose[curr] = True
        # print curr,

    for key in head.child:
        prinwords(curr+key, head.child[key])
        win[curr] = (win[curr] or not win[curr+key])
        lose[curr] = (lose[curr] or not lose[curr+key])

prinwords('', head)

if win[''] == False:
    print 'Second'
elif win[''] == True and lose[''] == True:
    print 'First'
elif win[''] == True and lose[''] == False:
    if k%2 == 0:
        print 'Second'
    else:
        print 'First'