Problem: Petya and Vasya arranged a game. The game runs by the following rules. Players have a directed graph consisting of n vertices and m edges. One of the vertices contains a chip. Initially the chip is located at vertex s. Players take turns moving the chip along some edge of the graph. Petya goes first. Player who can't move the chip loses. If the game lasts for 106 turns the draw is announced.

Vasya was performing big laboratory work in "Spelling and parts of speech" at night before the game, so he fell asleep at the very beginning of the game. Petya decided to take the advantage of this situation and make both Petya's and Vasya's moves.

Your task is to help Petya find out if he can win the game or at least draw a tie.

Input

The first line of input contain two integers n and m — the number of vertices and the number of edges in the graph (2 ≤ n ≤ 105, 0 ≤ m ≤ 2·105).

The next n lines contain the information about edges of the graph. i-th line (1 ≤ i ≤ n) contains nonnegative integer ci — number of vertices such that there is an edge from i to these vertices and ci distinct integers ai, j — indices of these vertices (1 ≤ ai, j ≤ n, ai, j ≠ i).

It is guaranteed that the total sum of ci equals to m.

The next line contains index of vertex s — the initial position of the chip (1 ≤ s ≤ n).

Output

If Petya can win print «Win» in the first line. In the next line print numbers v1, v2, ..., vk (1 ≤ k ≤ 106) — the sequence of vertices Petya should visit for the winning. Vertex v1 should coincide with s. For i = 1... k - 1 there should be an edge from vi to vi + 1 in the graph. There must be no possible move from vertex vk. The sequence should be such that Petya wins the game.

If Petya can't win but can draw a tie, print «Draw» in the only line. Otherwise print «Lose».

Examples

Input

5 6
2 2 3
2 4 5
1 4
1 5
0
1


Output

Win
1 2 4 5 


Input

3 2
1 3
1 1
0
2


Output

Lose


Input

2 2
1 2
1 1
1


Output

Draw

Note

In the first example the graph is the following:

<image>

Initially the chip is located at vertex 1. In the first move Petya moves the chip to vertex 2, after that he moves it to vertex 4 for Vasya. After that he moves to vertex 5. Now it is Vasya's turn and there is no possible move, so Petya wins.

In the second example the graph is the following:

<image>

Initially the chip is located at vertex 2. The only possible Petya's move is to go to vertex 1. After that he has to go to 3 for Vasya. Now it's Petya's turn but he has no possible move, so Petya loses.

In the third example the graph is the following:

<image>

Petya can't win, but he can move along the cycle, so the players will draw a tie.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("O3")
#pragma GCC optimize("O2")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (long long i = 0; i < v.size(); ++i) os << v[i] << " ";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  for (auto it : v) os << it << " ";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  os << v.first << " " << v.second;
  return os;
}
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
struct pair_hash {
  inline std::size_t operator()(
      const std::pair<long long, long long>& v) const {
    return v.first * 31 + v.second;
  }
};
const long long mod = 1e9 + 7;
long long takemod(long long a) { return ((a % mod) + mod) % mod; }
long long fast_exp(long long base, long long expo) {
  long long res = 1;
  while (expo > 0) {
    if (expo & 1) res = (res * base) % mod;
    base = (base * base) % mod;
    expo >>= 1;
  }
  return res;
}
long long modinv(long long a) { return takemod(fast_exp(takemod(a), mod - 2)); }
const long long N = 1e5 + 10;
vector<pair<long long, long long>> pg[N][2];
vector<long long> graph[N];
bool target[N];
bool doeswin = 0;
long long whatwin = -1;
bool vis[N][2];
long long vis2[N];
pair<long long, long long> parent[N][2];
bool istie = 0;
void dfs(long long a, long long p, long long pa = -1, long long pp = -1) {
  vis[a][p] = true;
  parent[a][p] = make_pair(pa, pp);
  if (target[a] == 1 and !p) {
    doeswin = 1;
    whatwin = a;
  }
  for (pair<long long, long long> px : pg[a][p]) {
    if (!vis[px.first][px.second]) {
      dfs(px.first, px.second, a, p);
    }
  }
}
void find_cycle(long long a) {
  vis2[a] = 1;
  for (long long i : graph[a]) {
    if (vis2[i] == 1) {
      istie = 1;
    } else if (!vis2[i]) {
      find_cycle(i);
    }
  }
  vis2[a] = 2;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  time_t t1, t2;
  t1 = clock();
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    long long c;
    cin >> c;
    if (!c) {
      ;
      target[i] = 1;
    }
    for (long long j = 0; j < c; j++) {
      long long x;
      cin >> x;
      graph[i].emplace_back(x);
      pg[i][0].emplace_back(make_pair(x, true));
      pg[i][1].emplace_back(make_pair(x, false));
    }
  }
  long long src;
  cin >> src;
  dfs(src, 1);
  if (doeswin) {
    long long x = whatwin, y = 0;
    vector<long long> v;
    v.emplace_back(x);
    while (parent[x][y] != make_pair(-1LL, -1LL)) {
      long long xn = parent[x][y].first;
      long long yn = parent[x][y].second;
      x = xn;
      y = yn;
      v.emplace_back(x);
    }
    reverse(v.begin(), v.end());
    cout << "Win" << '\n';
    cout << v;
    return 0;
  }
  find_cycle(src);
  if (istie) {
    cout << "Draw" << '\n';
  } else {
    cout << "Lose" << '\n';
  }
  t2 = clock();
  cerr << "time taken: " << t2 - t1;
  return 0;
}