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
#include <bits/stdc++.h>
using namespace std;
long long modulo(long long base, long long exponent, long long modulus);
long long choose(long long n, long long k);
long long inverse(long long a, long long m);
void build();
long long ncr(long long n, long long r);
const long long nax = 100010;
const long long LG = log2(nax) + 1;
vector<long long> fact(nax);
long long t[nax][26];
long long szt = 1;
long long root;
long long newT() { return szt++; }
void add(string s) {
  long long x = root;
  for (long long i = 0; i < s.size(); i++) {
    long long c = s[i] - 'a';
    if (t[x][c] == 0) t[x][c] = newT();
    x = t[x][c];
  }
}
long long win[nax], lose[nax];
void dfs(long long x) {
  bool leaf = true;
  for (long long i = 0; i < 26; i++) {
    if (t[x][i] != 0) {
      leaf = false;
      dfs(t[x][i]);
      win[x] |= !win[t[x][i]];
      lose[x] |= !lose[t[x][i]];
    }
  }
  if (leaf) {
    win[x] = false;
    lose[x] = true;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  root = newT();
  for (long long i = 0; i < n; i++) {
    string s;
    cin >> s;
    add(s);
  }
  dfs(root);
  if (win[root] && lose[root]) {
    cout << "First" << endl;
  } else if (!win[root]) {
    cout << "Second" << endl;
  } else if (win[root]) {
    if (k % 2 == 1)
      cout << "First" << endl;
    else
      cout << "Second" << endl;
  }
  return 0;
}
long long ncr(long long n, long long r) {
  if (r > n || r < 0 || n < 0) return 0;
  long long ans = fact[n];
  long long temp = (fact[n - r] * fact[r]) % 1000000007;
  ans = (ans * inverse(temp, 1000000007)) % 1000000007;
  return ans;
}
void build() {
  fact[0] = 1;
  for (long long i = 1; i < nax; i++) fact[i] = (fact[i - 1] * i) % 1000000007;
}
long long modulo(long long base, long long exponent, long long modulus) {
  if (modulus == 1) return 0;
  long long result = 1;
  base = base % modulus;
  while (exponent > 0) {
    if (exponent % 2 == 1) {
      result = (result * base) % modulus;
    }
    exponent = exponent >> 1;
    base = (base * base) % modulus;
  }
  return result;
}
long long choose(long long n, long long k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
void EE(long long a, long long b, long long &co1, long long &co2) {
  if (a % b == 0) {
    co1 = 0;
    co2 = 1;
    return;
  }
  EE(b, a % b, co1, co2);
  long long temp = co1;
  co1 = co2;
  co2 = temp - co2 * (a / b);
}
long long inverse(long long a, long long m) {
  long long x, y;
  EE(a, m, x, y);
  if (x < 0) x += m;
  return x;
}
