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 PYTHON3.
assistant
#Here is Python program to fix of at least one string from the group.

import math
import sys
from itertools import permutations
input = sys.stdin.readline

class Node:

    def __init__(self):
        self.children = [None]*26
        self.isEnd = False
        self.win = False
        self.lose = False

class Trie:

    def __init__(self):
        self.root = Node()

    def insert(self, key):
        cur = self.root
        for i in range(len(key)):
            if cur.children[ord(key[i])-ord('a')]==None:
                cur.children[ord(key[i])-ord('a')]=Node()
            cur = cur.children[ord(key[i])-ord('a')]
        
        cur.isEnd = True
    
    def search(self, key):
        cur = self.root
        for i in range(len(key)):
            if cur.children[ord(key[i])-ord('a')]==None:
                return False
            cur = cur.children[ord(key[i])-ord('a')]
        
        if cur!=None and cur.isEnd:
            return True

        return False
    
    def assignWin(self, cur):
        
        flag = True
        for i in range(26):
            if cur.children[i]!=None:
                flag=False
                self.assignWin(cur.children[i])
        
        if flag:
            cur.win=False
            cur.lose=True
        else:
            for i in range(26):
                if cur.children[i]!=None:
                    cur.win = cur.win or (not cur.children[i].win)
                    cur.lose = cur.lose or (not cur.children[i].lose)
       

if __name__=='__main__':
    
    t=Trie()

    n,k=map(int,input().split())
    for i in range(n):
        s=input()
        if s[-1]=="\n":
            s= s[:-1]
        t.insert(s)
    
    t.assignWin(t.root)
    
    if not t.root.win:
        print("Second")
    else:
        if t.root.lose:
            print("First")
        else:
            if k%2==1:
                print("First")
            else:
                print("Second")