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.

Difficulty: HARD
Tags: Dynamic programming

Graph algorithms

Graph traversal

Game theory
Skill types: Dynamic programming

Starter code:

Solutions:
import sys
data = sys.stdin.readlines()
(n, m) = map(int, data[0].split())
g = {}
for (i, line) in enumerate(data[1:-1], 1):
	g[i] = list(map(int, line.split()[1:]))
mk = {}
start = int(data[-1])
queue = [(start, 0, -1, 1)]
cycle = False
while len(queue) > 0:
	(v, player, prev, color) = queue.pop()
	if color == 2:
		mk[v, player] = (prev, 2)
		continue
	if mk.get((v, player), None):
		if mk[v, player][1] == 1:
			cycle = True
		continue
	mk[v, player] = (prev, 1)
	queue.append((v, player, prev, 2))
	for w in g[v]:
		queue.append((w, 1 - player, v, 1))
sol = None
for v in range(1, n + 1):
	if len(g[v]) == 0 and mk.get((v, 1), None):
		sol = v
		break
if sol:
	path = [sol]
	cur = (sol, 1)
	while cur != (start, 0):
		cur = (mk.get(cur)[0], 1 - cur[1])
		path.append(cur[0])
	print('Win')
	print(*path[::-1])
elif cycle:
	print('Draw')
else:
	print('Lose')


(n, m) = [int(x) for x in input().split()]
edges = [set()]
drain = set()
for i in range(n):
	nums = [int(x) for x in input().split()]
	if nums[0] == 0:
		drain.add(i + 1)
	edges.append(set(nums[1:]))
start = int(input())
nodes = {start}
visited = {}
while len(nodes) > 0:
	curr = nodes.pop()
	for v in edges[abs(curr)]:
		if curr > 0:
			v = -v
		if visited.get(v) is None:
			nodes.add(v)
			visited[v] = curr
ok = False
for d in drain:
	if -d in visited:
		print('Win')
		v = -d
		l = []
		while v != start:
			l.append(abs(v))
			v = visited[v]
		l.append(start)
		print(' '.join((str(x) for x in l[::-1])))
		ok = True
		break
if not ok:
	stack = [start]
	in_stack = {start}
	visited = {start}
	while len(stack) > 0:
		curr = stack[-1]
		if len(edges[curr]) == 0:
			stack.pop()
			in_stack.remove(curr)
			continue
		v = edges[curr].pop()
		if v not in visited:
			visited.add(v)
			stack.append(v)
			in_stack.add(v)
		elif v in in_stack:
			print('Draw')
			ok = True
			break
	if not ok:
		print('Lose')


(n, m) = map(int, input().split())
g = [[] for i in range(n)]
fs = set()
for i in range(n):
	a = list(map(int, input().split()))
	c = a[0]
	if c == 0:
		fs.add(i)
		continue
	for j in range(1, c + 1):
		g[i].append(a[j] - 1)
s = int(input()) - 1
prev0 = [None for i in range(n)]
prev1 = [None for i in range(n)]
vis0 = [0 for i in range(n)]
vis0[s] = 1
vis1 = [0 for i in range(n)]
q = [(s, 0)]
ans = None
draw = False
while len(q) > 0:
	(v, c) = q[0]
	del q[0]
	for u in g[v]:
		if c == 0:
			if vis1[u] == 0:
				vis1[u] = 1
				q.append((u, 1))
				prev1[u] = v
			if u in fs:
				ans = u
				break
		elif c == 1:
			if vis0[u] == 0:
				vis0[u] = 1
				q.append((u, 0))
				prev0[u] = v
	if ans is not None:
		break
if ans is None:
	q = [s]
	vis = [0 for i in range(n)]
	vis[s] = 1
	nxt = [0 for i in range(n)]
	while len(q) > 0:
		v = q[-1]
		if nxt[v] < len(g[v]):
			u = g[v][nxt[v]]
			if vis[u] == 1:
				print('Draw')
				exit()
			elif vis[u] == 0:
				vis[u] = 1
				q.append(u)
			nxt[v] += 1
		else:
			vis[v] = 2
			del q[-1]
	print('Lose')
	exit()
arn = []
nxt = ans
while nxt is not None:
	arn.append(nxt)
	if len(arn) % 2 == 1:
		nxt = prev1[nxt]
	else:
		nxt = prev0[nxt]
print('Win')
arn = list(reversed(arn))
print(' '.join([str(i + 1) for i in arn]))


(n, m) = list(map(int, input().split()))
from collections import defaultdict
graph = defaultdict(list)
for i in range(1, n + 1):
	neighbours = list(map(int, input().split()))[1:]
	graph[i, 0] = [(x, 1) for x in neighbours]
	graph[i, 1] = [(x, 0) for x in neighbours]
start = int(input())
terminal_states = []
for x in graph:
	if len(graph[x]) == 0 and x[1] == 1:
		terminal_states.append(x[0])
import sys
stack = [(start, 0)]
parents = {}
visited = {}
visited[start, 0] = True
parents[start, 0] = None
while len(stack) != 0:
	current = stack.pop()
	for neighbour in graph[current]:
		if neighbour not in visited:
			stack.append(neighbour)
			visited[neighbour] = True
			parents[neighbour] = current
for terminal in terminal_states:
	if (terminal, 1) in visited:
		print('Win')
		solution = []
		current = (terminal, 1)
		while current != None:
			solution.append(current[0])
			current = parents[current]
		print(' '.join(list(map(str, solution[::-1]))))
		sys.exit()
from queue import Queue
ingoing_graph = defaultdict(int)
for node in visited:
	ingoing_graph[node] += 0
	for neighbour in graph[node]:
		if neighbour in visited:
			ingoing_graph[neighbour] += 1
zero_indegree_nodes = []
deleted = {}
for node in ingoing_graph:
	if node in visited and ingoing_graph[node] == 0:
		zero_indegree_nodes.append(node)
count = 0
while zero_indegree_nodes:
	current = zero_indegree_nodes.pop()
	count += 1
	for neighbour in graph[current]:
		if neighbour not in deleted:
			ingoing_graph[neighbour] -= 1
			if ingoing_graph[neighbour] == 0:
				zero_indegree_nodes.append(neighbour)
				deleted[neighbour] = True
if count < len(visited):
	print('Draw')
else:
	print('Lose')


import collections

def do():
	(nodes, edges) = map(int, input().split(' '))
	outd = [0] * (nodes + 1)
	g = collections.defaultdict(list)
	for i in range(1, nodes + 1):
		tmp = [int(c) for c in input().split(' ')]
		outd[i] = tmp[0]
		g[i] = tmp[1:]
	dp = [[0] * 2 for _ in range(nodes + 1)]
	pre = [[0] * 2 for _ in range(nodes + 1)]
	st = int(input())
	mask = [0] * (nodes + 1)

	def dfs(entry):
		has_loop = False
		stack = [[entry, 0, True]]
		dp[entry][0] = 1
		while stack:
			(cur, step, first) = stack.pop()
			if first:
				mask[cur] = 1
				stack.append([cur, -1, False])
				for nei in g[cur]:
					if mask[nei]:
						has_loop = True
					if dp[nei][step ^ 1]:
						continue
					pre[nei][step ^ 1] = cur
					dp[nei][step ^ 1] = 1
					stack.append([nei, step ^ 1, first])
			else:
				mask[cur] = 0
		return has_loop
	has_loop = dfs(st)
	for i in range(1, nodes + 1):
		if outd[i] == 0:
			if dp[i][1]:
				print('Win')
				res = []
				cur = i
				step = 1
				while cur != st or step:
					res.append(cur)
					cur = pre[cur][step]
					step ^= 1
				res.append(st)
				print(' '.join((str(c) for c in res[::-1])))
				return
	res = 'Draw' if has_loop else 'Lose'
	print(res)
do()