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
# -*- coding: utf-8 -*-
def inpl(): return tuple(map(int, input().split()))
INF = float("inf")
V, E, r = inpl()

edges = [inpl() for _ in range(E)]

def belman_ford(V, edges, r):
    "If glaph contains negative cycles then return True"
    D = [INF]*(V)
    D[r] = 0
    if V == 1:
        return False, [0]
    
    for _ in range(V-1):
        update = False
        for s, t, d in edges:
            if D[t] > D[s] + d:
                D[t] = D[s] + d
                update = True
        if not update:
            break
    else:
        return True, D
    return False, D
    
NEGATIVE, D = belman_ford(V, edges, r)

if NEGATIVE:
    print("NEGATIVE CYCLE")
else:
    for d in D:
        print("INF") if d == INF else print(d)
