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
V, E, r = map(int, input().split())
inf = 2**32
e = []
for i in range(E):
    s, t, d = map(int, input().split())
    e.append([s, t, d])

# 始点からの距離初期化
distance = [inf]*V
distance[r] = 0

'''
ベルマンフォード法による単一最短経路
負の重みがある場合でも大丈夫だが、負の重みがない場合はダイクストラのほうがよい
O(V*E)
'''


def bellmanFord():
    for v in range(V):
        updated = False
        for s, t, d in e:
            if distance[s] != inf and (distance[s]+d) < distance[t]:
                distance[t] = distance[s]+d
                updated = True
        if v == (V-1) and updated:
            return False
    return True


b = bellmanFord()
if b:
    for v in range(V):
        print(distance[v] if distance[v] != inf else "INF")
else:
    print("NEGATIVE CYCLE")

