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
class exist_negative_cycle(Exception):
    pass
def bellman_ford(g, size, start=0):
    d = [float('inf')] * size
    d[start] = 0
   
    for _ in range(size):
        for u in g:
            for length, v in g[u]:
                if d[v] > d[u] + length:
                    d[v] = d[u] + length

    for u in g:
        for length, v in g[u]:
            if d[v] > d[u] + length:
                raise exist_negative_cycle    
    return d
from sys import stdin
readline = stdin.readline
import math
from collections import defaultdict

vertex, edge, r = map(int, readline().split())

g = defaultdict(list)
for i in range(edge):
    s, t, length = map(int, readline().split())
    g[s].append((length, t))
try:
    d = bellman_ford(g, vertex, r)
    for di in d:
        print('INF' if math.isinf(di) else di)
except exist_negative_cycle:
    print('NEGATIVE CYCLE')