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:
import sys
fin = sys.stdin.readline


MAX_NUM = float('inf')

def initialize_int(adj, s):
    """
    assume that all vertex in {0, 1, 2, ..., N - 1}
    """
    N = len(adj)
    d = [MAX_NUM] * N
    d[s] = 0
    parent = [None] * N
    return d, parent


def relax(d, w, source_v, target_v, parent):
    new_cost = d[source_v] + w[source_v][target_v]
    if new_cost < d[target_v]:
        d[target_v] = new_cost
        parent[target_v] = source_v


# Even if a negative weight exists, bellman-ford can find shortest paths.
# it also detects negative cycles,
# but not every edges that consist the negative cycles is stored.
# time complexity: O(VE + (V^2))
def bellman_ford(adj, w, s):
    d, parent = initialize_int(adj, s)

    # calculate shortest paths
    for _ in range(len(adj)):
        for vertex in range(len(adj)):
            for neighbor in adj[vertex]:
                relax(d, w, vertex, neighbor, parent)

    # detects negative cycles if exist
    negative_cycle_edges = set()
    for vertex in range(len(adj)):
        for neighbor in adj[vertex]:
            if d[neighbor] > d[vertex] + w[vertex][neighbor]:
                negative_cycle_edges.add((vertex, neighbor))
    return d, parent, negative_cycle_edges



V, E, r = [int(elem) for elem in fin().split()]
adj = [[] for _ in range(V)]
w = [[None] * V for _ in range(V)]
for _ in range(E):
    s, t, d = [int(elem) for elem in fin().split()]
    adj[s].append(t)
    w[s][t] = d

d, parent, negative_cycle_edges = bellman_ford(adj, w, r)
if len(negative_cycle_edges) > 0:
    print("NEGATIVE CYCLE")
else:
    for i in range(V):
        print("INF") if d[i] == MAX_NUM else print(d[i])