Problem: 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

Solution (in PYTHON3):

Solution:
from collections import defaultdict

INF = 999999999999999999999
NEGATIVE_CYCLE = -1

def bellman_ford(G, n, r):
  d = {}
  for i in range(n):
    d[i] = INF
  d[r] = 0
  for i in range(n-1):
    d = relax(G, d)
  for u in G:
    for v in G[u]:
      if d[u] == INF and d[v] == INF:
        continue
      if d[u] + G[u][v] < d[v]:
        return NEGATIVE_CYCLE
  return d

def relax(G, d):
  for u in G:
    if d[u] == INF:
      continue
    for v in G[u]:
      if d[u] + G[u][v] < d[v]:
        d[v] = d[u] + G[u][v]
  return d


G = {}

n, m, r = [int(v) for v in input().strip().split(' ')]
for i in range(n):
  G[i] = {}
for _ in range(m):
  s, t, c = [int(v) for v in input().strip().split(' ')]
  G[s][t] = c

d = bellman_ford(G, n, r)

if d == NEGATIVE_CYCLE:
  print("NEGATIVE CYCLE")
else:
  for i in range(n):
    if d[i] == INF:
      print("INF")
    else:
      print(d[i])