Problem: 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

Solution (in PYTHON3):

Solution:
from heapq import heappop,heappush,heapify

class Graph():
    def __init__(self,n,edge,directed=False):
        self.n = n
        self.graph = [[] for _ in range(n)]
        for e in edge:
            self.graph[e[0]].append((e[1],e[2]))
            if not directed:
                self.graph[e[1]].append((e[0],e[2]))
    def bellman_ford(self,s,INF=10**18):
        dist = [INF for _ in range(self.n)]
        dist[s] = 0
        update = True
        for i in range(self.n):
            update = False
            for node, g in enumerate(self.graph):
                for adj,adjcost in g:
                    if dist[node] != INF and dist[node]+adjcost < dist[adj]:
                        dist[adj] = dist[node]+adjcost
                        update = True
            if not update:
                return dist
        return False

INF = 10**18

import sys
input = sys.stdin.readline

V,E,r = map(int,input().split())
edge = [tuple(map(int,input().split())) for _ in range(E)]

g = Graph(V,edge,True)
D = g.bellman_ford(r,INF)

if D:
    for i in range(V):
        if D[i] == INF: D[i] = 'INF'
    print('\n'.join(map(str,D)))

else:
    print('NEGATIVE CYCLE')