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
def m ... What is 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.
ain():
    n, k = [int(x) for x in input().split()]
    S = [tuple(ord(c)-97 for c in input()) for _ in range(n)]

    @bootstrap
    def winners(S,i):
        C = [[] for _ in range(26)]
        S = [s for s in S if len(s)!=i]
        if not S:
            yield False

        for s in S:
            C[s[i]].append(s)

        for c in C:
            if c and not (yield winners(c,i+1)):
                yield True
        yield False
    
    @bootstrap
    def losers(S,i):
        C = [[] for _ in range(26)]
        S = [s for s in S if len(s)!=i]
        if not S:
            yield True

        for s in S:
            C[s[i]].append(s)

        for c in C:
            if c and not (yield losers(c,i+1)):
                yield True
        yield False

    a = winners(S,0)
    b = losers(S,0)
    #print a,b

    if a and b:
        print 'First'
    elif (not a) and (not b):
        print 'Second'
    elif b:
        print 'Second'
    else:
        print 'First' if k%2 == 1 else 'Second'
        



######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s

   ...