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
from collections import defaultdict

def bellman_ford(V, E, s):
    '''
    Input:
        V = list(range(N))
        E = {i: defaultdict(int) for i in range(N)}
        s (in V) : start vertex
    Output:
        dist: the list of the minimum cost from s to arbitrary vertex v in the given graph
    '''
    INF = float('inf')
    N = len(V)
    dist = [INF]*N
    prev = [-1]*N
    dist[s] = 0
    for n_iterations in range(N):
        updated = False
        for v in V:
            d = dist[v]
            if d != INF:
                for u, c in E[v].items():
                    temp = d + c
                    if dist[u] > temp:
                        updated = True
                        dist[u] = temp
                        prev[u] = v
        if not updated: # the minimum costs are already stored in dist
            break
    else: # the above loop is not broken if and only if there exists a negative cycle
        return None
    return dist

INF = float('inf')
N, M, r = map(int, input().split())
E = {i: defaultdict(int) for i in range(N)}
for _ in range(M):
    a, b, c = map(int, input().split())
    E[a][b] = c
V = list(range(N))
dist = bellman_ford(V, E, r)
if dist is None:
    print('NEGATIVE CYCLE')
else:
    for d in dist:
        if d == INF: print('INF')
        else: print(d)
