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
from collections import namedtuple, deque


INF = sys.maxsize
Edge = namedtuple('Edge', 's t d')

V, E, r = map(int, input().split())
edges = []

for _ in range(E):
  s, t, d = map(int, input().split())
  edges.append(Edge(s, t, d))

def solve():
  costs = [INF] * V
  costs[r] = 0
  
  for i in range(V):      
    for edge in edges:
      if costs[edge.s] != INF and costs[edge.t] > costs[edge.s] + edge.d:
        if i == V - 1:
          print('NEGATIVE CYCLE')
          return
          
        costs[edge.t] = costs[edge.s] + edge.d        
      
  for x in costs:
    print(x if x != INF else 'INF')

solve()

