455_B. A Lot of Games
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
```python
"""
Codeforces Contest 260 Div 1 Problem B

Author  : chaotic_iak
Language: Python 3.3.4
"""

def main():
    n,k = read()
    s = set()
    for i in range(n): s.add(read(0))
    s = list(s)
    s.sort()
    s = treeify(s)
    res = solve(s)
    if res == 0: # neither: second player win
        print("Second")
    if res == 1: # odd: first player win if k is odd
        print("First" if k % 2 else "Second")
    if res == 2: # even: second player win
        print("Second")
    if res == 3: # both: first player win
        print("First")

def treeify(s):
    res = [[] for _ in range(26)]
    for i in s:
        if i: res[ord(i[0]) - 97].append(i[1:])
    fin = []
    for i in range(26):
        if res[i]: fin.append(treeify(res[i]))
    return fin

def solve(s, parity=2):
    for i in range(len(s)):
        if isinstance(s[i], list): s[i] = solve(s[i], 3-parity)
    if not s: return parity # no possible move: current parity
    if 0 in s: return 3 # any neither: both
    if 1 in s and 2 in s: return 3 # any odd and any even: both
    if 1 in s: return 1 # any odd: odd
    if 2 in s: return 2 # any even: even
    return 0 # all both: neither

################################### NON-SOLUTION STUFF BELOW

def read(mode=2):
    # 0: String
    # 1: List of strings
    # 2: List of integers
    inputs = input().strip()
    if mode == 0: return inputs
    if mode == 1: return inputs.split()
    if mode == 2: return map(int, inputs.split())

def write(s="\n"):
    if isinstance(s, list): s = " ".join(map(str, s))
    s = str(s)
    print(s, end="")

main()
```