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 = float('inf')

def Bellmanford(n, edges, r): #r: ?§????
    d = [INF] * n
    d[r] = 0
    
    for i in range(n):
        
        for (u, v, c) in edges:
            if d[u] != INF and d[u] + c < d[v]:
                d[v] = d[u] + c
                if i == n-1:
                    return ['NEGATIVE CYCLE']
    
    d = [di if di < INF else 'INF' for di in d]
    return d
    

V, E, R = map(int, sys.stdin.readline().split())
Edges = [None] * E

for i in range(E):
    si, ti, di = map(int, sys.stdin.readline().split())
    Edges[i] = (si, ti, di)
    
ans = Bellmanford(V, Edges, R)

print(*ans, sep = '\n')