Recently Irina arrived to one of the most famous cities of Berland — the Berlatov city. There are n showplaces in the city, numbered from 1 to n, and some of them are connected by one-directional roads. The roads in Berlatov are designed in a way such that there are no cyclic routes between showplaces.

Initially Irina stands at the showplace 1, and the endpoint of her journey is the showplace n. Naturally, Irina wants to visit as much showplaces as she can during her journey. However, Irina's stay in Berlatov is limited and she can't be there for more than T time units.

Help Irina determine how many showplaces she may visit during her journey from showplace 1 to showplace n within a time not exceeding T. It is guaranteed that there is at least one route from showplace 1 to showplace n such that Irina will spend no more than T time units passing it.


-----Input-----

The first line of the input contains three integers n, m and T (2 ≤ n ≤ 5000,  1 ≤ m ≤ 5000,  1 ≤ T ≤ 10^9) — the number of showplaces, the number of roads between them and the time of Irina's stay in Berlatov respectively.

The next m lines describes roads in Berlatov. i-th of them contains 3 integers u_{i}, v_{i}, t_{i} (1 ≤ u_{i}, v_{i} ≤ n, u_{i} ≠ v_{i}, 1 ≤ t_{i} ≤ 10^9), meaning that there is a road starting from showplace u_{i} and leading to showplace v_{i}, and Irina spends t_{i} time units to pass it. It is guaranteed that the roads do not form cyclic routes.

It is guaranteed, that there is at most one road between each pair of showplaces.


-----Output-----

Print the single integer k (2 ≤ k ≤ n) — the maximum number of showplaces that Irina can visit during her journey from showplace 1 to showplace n within time not exceeding T, in the first line.

Print k distinct integers in the second line — indices of showplaces that Irina will visit on her route, in the order of encountering them.

If there are multiple answers, print any of them.


-----Examples-----
Input
4 3 13
1 2 5
2 3 7
2 4 8

Output
3
1 2 4 

Input
6 6 7
1 2 2
1 3 3
3 6 3
2 4 2
4 6 2
6 5 1

Output
4
1 2 4 6 

Input
5 5 6
1 3 3
3 5 3
1 2 2
2 4 3
4 5 2

Output
3
1 3 5

Difficulty: MEDIUM_HARD
Tags: Dynamic programming

Graph algorithms
Skill types: Dynamic programming

Starter code:

Solutions:
(n, m, T) = map(int, input().split())
graph_a = [[] for _ in range(n + 1)]
graph_b = [[] for _ in range(n + 1)]
double_graph_a = [[0 for _ in range(n + 1)] for _ in range(n + 1)]
double_graph_b = [[0 for _ in range(n + 1)] for _ in range(n + 1)]
for i in range(m):
	(u, v, t) = map(int, input().split())
	graph_a[v].append(u)
	graph_b[v].append(t)
	if u == 1:
		double_graph_a[v][2] = t
		double_graph_b[v][2] = 1
next_n = n + 1
for c in range(3, next_n):
	for v in range(2, next_n):
		for (i, nx) in enumerate(graph_a[v]):
			if double_graph_a[nx][c - 1]:
				dist = double_graph_a[nx][c - 1] + graph_b[v][i]
				if dist <= T and (not double_graph_a[v][c] or dist < double_graph_a[v][c]):
					double_graph_a[v][c] = dist
					double_graph_b[v][c] = nx
for i in range(n, 0, -1):
	if double_graph_b[n][i]:
		break
res = [n]
while double_graph_b[res[-1]][i] != 1:
	res.append(double_graph_b[res[-1]][i])
	i -= 1
res += [1]
print(len(res))
print(' '.join(map(str, res[::-1])))


(n, m, T) = map(int, input().split())
graph_a = [[] for _ in range(n + 1)]
graph_b = [[] for _ in range(n + 1)]
double_graph_a = [[0 for _ in range(n + 1)] for _ in range(n + 1)]
double_graph_b = [[0 for _ in range(n + 1)] for _ in range(n + 1)]
for i in range(m):
	(u, v, t) = map(int, input().split())
	graph_a[v].append(u)
	graph_b[v].append(t)
	if u == 1:
		double_graph_a[v][2] = t
		double_graph_b[v][2] = 1
next_n = n + 1
for i in range(3, next_n):
	for j in range(2, next_n):
		for (a, b) in enumerate(graph_a[j]):
			if double_graph_a[b][i - 1]:
				dist = double_graph_a[b][i - 1] + graph_b[j][a]
				if dist <= T and (not double_graph_a[j][i] or dist < double_graph_a[j][i]):
					double_graph_a[j][i] = dist
					double_graph_b[j][i] = b
start = n
stop = 0
step = -1
for i in range(start, stop, step):
	if double_graph_b[start][i]:
		break
res = [n]
while double_graph_b[res[-1]][i] != 1:
	res.append(double_graph_b[res[-1]][i])
	i -= 1
res += [1]
print(len(res))
print(' '.join(map(str, res[::-1])))


(n, m, t) = map(int, input().split())
edge = {i: {} for i in range(n)}
income = [0 for i in range(n)]
for i in range(m):
	(u, v, ti) = map(int, input().split())
	edge[v - 1][u - 1] = ti
	income[u - 1] += 1
stat = [{} for _ in range(n)]
stat[n - 1] = {1: (0, -1)}
queue = [n - 1]
first = 0
last = 1
for i in range(n - 2, 0, -1):
	if income[i] == 0:
		queue.append(i)
		last += 1
while first < last:
	v = queue[first]
	first += 1
	for u in edge[v].keys():
		income[u] -= 1
		for vis in stat[v].keys():
			cost = stat[v][vis][0] + edge[v][u]
			ucost = stat[u].get(vis + 1, (t + 1, -1))[0]
			if ucost > cost:
				stat[u][vis + 1] = (cost, v)
		if income[u] <= 0:
			queue.append(u)
			last += 1
res = max(stat[0].keys())
print(res)
path = []
curr = 0
path.append(curr + 1)
while stat[curr][res][1] >= 0:
	curr = stat[curr][res][1]
	path.append(curr + 1)
	res -= 1
print(' '.join(map(str, path)))


(n, m, T) = map(int, input().split())
adj = [[] for _ in range(n + 1)]
adw = [[] for _ in range(n + 1)]
dp = [[0 for _ in range(n + 1)] for _ in range(n + 1)]
pv = [[0 for _ in range(n + 1)] for _ in range(n + 1)]
for i in range(m):
	(a, b, t) = map(int, input().split())
	adj[b].append(a)
	adw[b].append(t)
	if a == 1:
		dp[b][2] = t
		pv[b][2] = 1
for c in range(3, n + 1):
	for v in range(2, n + 1):
		for (i, nx) in enumerate(adj[v]):
			if dp[nx][c - 1]:
				dist = dp[nx][c - 1] + adw[v][i]
				if dist <= T and (not dp[v][c] or dist < dp[v][c]):
					dp[v][c] = dist
					pv[v][c] = nx
for i in range(n, 0, -1):
	if pv[n][i]:
		break
res = [n]
while pv[res[-1]][i] != 1:
	res.append(pv[res[-1]][i])
	i -= 1
res += [1]
print(len(res))
print(' '.join(map(str, res[::-1])))


(n, m, T) = map(int, input().split())
adj = [[] for _ in range(n + 1)]
ad_w = [[] for _ in range(n + 1)]
dp = [[0 for _ in range(n + 1)] for _ in range(n + 1)]
pv = [[0 for _ in range(n + 1)] for _ in range(n + 1)]
for i in range(m):
	(a, b, t) = map(int, input().split())
	adj[b].append(a)
	ad_w[b].append(t)
	if a == 1:
		dp[b][2] = t
		pv[b][2] = 1
for c in range(3, n + 1):
	for v in range(2, n + 1):
		for (i, nx) in enumerate(adj[v]):
			if dp[nx][c - 1]:
				newdistance = dp[nx][c - 1] + ad_w[v][i]
				if newdistance <= T and (not dp[v][c] or newdistance < dp[v][c]):
					dp[v][c] = newdistance
					pv[v][c] = nx
for last in range(n, 0, -1):
	if pv[n][last]:
		break
path = [n]
while pv[path[-1]][last] != 1:
	path.append(pv[path[-1]][last])
	last -= 1
path.append(1)
path.reverse()
print(len(path))
print(' '.join(map(str, path)))


from sys import stdin, stdout

class LLNode:

	def __init__(self, value=None, nextNode=None):
		(self.value, self.nextNode) = (value, nextNode)

class LinkedList:

	def __init__(self, rootNode=None):
		self.rootNode = rootNode

	def FindBestPath(self, wayTime):
		cNode = self.rootNode
		while cNode != None:
			if cNode.value.wayTime <= wayTime:
				break
			cNode = cNode.nextNode
		return cNode

	def FindMaxPreNode(self, path, startNode):
		if path.wayLen > startNode.value.wayLen:
			return None
		preNode = startNode
		while preNode.nextNode != None:
			if path.wayLen > preNode.nextNode.value.wayLen:
				break
			preNode = preNode.nextNode
		return preNode

	def FindMaxNextNode(self, path, startNode):
		cNode = startNode
		while cNode != None:
			if cNode.value.wayTime < path.wayTime:
				break
			cNode = cNode.nextNode
		return cNode

	def RawInsert(self, path):
		if path == None:
			return
		if self.rootNode == None:
			self.rootNode = LLNode(path, None)
			return
		preNode = self.FindMaxPreNode(path, self.rootNode)
		preNode.nextNode = LLNode(path, None)

	def Insert(self, path):
		if path == None:
			return False
		if self.rootNode == None:
			self.rootNode = LLNode(path, None)
			return True
		preNode = self.FindMaxPreNode(path, self.rootNode)
		node = None
		if preNode == None:
			node = LLNode(path, self.rootNode)
			self.rootNode = node
		elif path.wayTime < preNode.value.wayTime:
			if path.wayLen == preNode.value.wayLen:
				preNode.value = path
				node = preNode
			else:
				node = LLNode(path, preNode.nextNode)
				preNode.nextNode = node
		else:
			return False
		if node.nextNode != None:
			nextNode = self.FindMaxNextNode(path, node.nextNode)
			node.nextNode = nextNode
		return True
cities = []
memo = []
T = 0

class Node:

	def __init__(self, number, wayLen, wayTime, preNode):
		(self.number, self.wayLen, self.wayTime, self.preNode) = (number, wayLen, wayTime, preNode)

class Path:

	def __init__(self, wayLen, wayTime, nextNode):
		(self.wayLen, self.wayTime, self.nextNode) = (wayLen, wayTime, nextNode)

def MemoizeWay(endNode, tailLen, tailTime):
	global memo
	cNode = endNode
	while cNode.preNode != None:
		tailLen += 1
		tailTime += cities[cNode.preNode.number][cNode.number]
		b = memo[cNode.preNode.number].Insert(Path(tailLen, tailTime, cNode.number))
		cNode = cNode.preNode

def main():
	global T, cities
	(n, m, T) = [int(s) for s in stdin.readline().split()]
	cities = [None] * (n + 1)
	for i in range(m):
		(u, v, t) = [int(s) for s in stdin.readline().split()]
		if cities[u] == None:
			cities[u] = {}
		if u != n:
			cities[u].update({v: t})
	global memo
	memo = [LinkedList() for i in range(n + 1)]
	memo[n].Insert(Path(1, 0, None))
	stack = []
	stack.append(Node(1, 1, 0, None))
	while len(stack) > 0:
		city = stack.pop()
		if city.number == n:
			MemoizeWay(city, 1, 0)
			continue
		if memo[city.number].rootNode != None:
			cNode = memo[city.number].rootNode
			while cNode != None:
				MemoizeWay(city, cNode.value.wayLen, cNode.value.wayTime)
				cNode = cNode.nextNode
			continue
		if cities[city.number] != None:
			for subCity in cities[city.number]:
				stack.append(Node(subCity, city.wayLen + 1, city.wayTime + cities[city.number][subCity], city))
	cPath = memo[1].FindBestPath(T).value
	stdout.write('{}\n'.format(cPath.wayLen))
	stdout.write('{} '.format(1))
	while cPath.nextNode != None:
		stdout.write('{} '.format(cPath.nextNode))
		cPath = memo[cPath.nextNode].FindBestPath(cPath.wayTime).value
main()


import collections
import time
import os
import sys
import bisect
import heapq
from typing import List

def solve(N, M, T, edges):
	g = collections.defaultdict(list)
	for (u, v, t) in edges:
		g[u].append((v, t))
	dp = [[T + 1 for _ in range(N + 1)] for _ in range(N + 1)]
	pre = [[0 for _ in range(N + 1)] for _ in range(N + 1)]
	dp[1][1] = 0
	pre[1][1] = 0
	q = [(1, 1, 0)]
	while q:
		nq = []
		for (city, dist, pcost) in q:
			dist += 1
			for (dest, ncost) in g[city]:
				cost = pcost + ncost
				if cost <= T and dp[dest][dist] > cost:
					dp[dest][dist] = cost
					pre[dest][dist] = city
					nq.append((dest, dist, cost))
		q = nq
	ans = max([i for i in range(N, -1, -1) if dp[N][i] <= T])
	print(ans)
	path = []
	k = N
	l = ans
	while k:
		path.append(k)
		k = pre[k][l]
		l -= 1
	print(' '.join(map(str, path[::-1])))
(N, M, T) = map(int, input().split())
edges = []
for i in range(M):
	(u, v, t) = map(int, input().split())
	edges.append((u, v, t))
solve(N, M, T, edges)