Constraints

* 1 ≤ |V| ≤ 1000
* 0 ≤ |E| ≤ 2000
* -10000 ≤ di ≤ 10000
* There are no parallel edges
* There are no self-loops

Input

An edge-weighted graph G (V, E) and the source r.


|V| |E| r
s0 t0 d0
s1 t1 d1
:
s|E|-1 t|E|-1 d|E|-1


|V| is the number of vertices and |E| is the number of edges in G. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively. r is the source of the graph.

si and ti represent source and target vertices of i-th edge (directed) and di represents the cost of the i-th edge.

Output

If the graph contains a negative cycle (a cycle whose sum of edge costs is a negative value) which is reachable from the source r, print


NEGATIVE CYCLE


in a line.

Otherwise, print


c0
c1
:
c|V|-1


The output consists of |V| lines. Print the cost of the shortest path from the source r to each vertex 0, 1, ... |V|-1 in order. If there is no path from the source to a vertex, print "INF".

Examples

Input

4 5 0
0 1 2
0 2 3
1 2 -5
1 3 1
2 3 2


Output

0
2
-3
-1


Input

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


Output

NEGATIVE CYCLE


Input

4 5 1
0 1 2
0 2 3
1 2 -5
1 3 1
2 3 2


Output

INF
0
-5
-3
import sys
INF = 1000000000

class Edge:
	def __init__(self, to, cost):
		self.to = to
		self.cost = cost


def bellman_ford(graph, start):
	V = len(graph)
	d = [INF for _ in xrange(V)]
	d[start] = 0
	
	for _ in xrange(V-1):
		for i, node in enumerate(graph):
			for edge in node:
				if d[i] != INF and d[i] + edge.cost < d[edge.to]:
					d[edge.to] = d[i] + edge.cost

	for i, node in enumerate(graph):
		for edge in node:
			if d[i] != INF and d[i] + edge.cost < d[edge.to]:
				print "NEGATIVE CYCLE"
				sys.exit()

	return d

def main():
	V, E, r = map(int, raw_input().split())

	graph = [[] for _ in xrange(V)]
	for i in xrange(E):
		_s, _t, _c = map(int, raw_input().split())
		graph[_s].append(Edge(_t, _c))


	distance = bellman_ford(graph, r)


	for d in distance:
		if d == INF:
			print "INF"
		else:
			print d

if __name__ == '__main__':
	main()