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 heapq import heappush, heappop


def bellman_ford(V, Edges, start):
    dis = {}
    
    for i in range(V): dis[i] = (1 << 30)
    
    dis[start] = 0
    
    for i in range(V):
        for a, b, w in Edges:
            if dis[a] != (1 << 30) and dis[a] + w < dis[b]:
                dis[b] = dis[a] + w
    
    for a, b, w in Edges:
        if dis[a] != (1 << 30) and dis[a] + w < dis[b]:
            return ('NEGATIVE CYCLE', dis)
    
    return ('No NEGATIVE CYCLE', dis)
    
    
if __name__ == '__main__':
    line = sys.stdin.readline()
    V, E, start = line.split()
    V, E, start = int(V), int(E), int(start)
    Es = []
    for i in range(E):
        line = sys.stdin.readline()
        a, b, w = line.split()
        a, b, w = int(a), int(b), int(w)
        Es.append((a,b,w))
        
    judge, dis = bellman_ford(V, Es, start)
    
    if judge == 'NEGATIVE CYCLE':
        sys.stdout.write(judge + '\n')
    else:
        for i in range(V):
            if dis[i] == (1 << 30):
                sys.stdout.write('INF\n')
            else:
                sys.stdout.write(str(dis[i]) + '\n')