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 deque
N, M, r = map(int, input().split())
G = [[] for i in range(N)]
for i in range(M):
    s, t, d = map(int, input().split())
    G[s].append((t, d))

INF = 10**18
def SPFA(N, G, s):
    dist = [INF] * N
    cont = [0] * N
    cnts = [0]* N

    dist[s] = 0
    cont[s] = 1
    cnts[s] += 1
    que = deque([s])
    su = s; num = 1
    while que:
        v = que.popleft()
        cont[v] = 0
        d = dist[v]
        su -= d
        num -= 1
        for w, c in G[v]:
            r = d + c
            if r < dist[w]:
                if not cont[w]:
                    cnts[w] += 1
                    if N <= cnts[w]:
                        return None
                    if que and r < que[0]:
                        que.appendleft(w)
                    else:
                        que.append(w)
                    cont[w] = 1
                    su += r
                    num += 1
                    dist[w] = r
                    while su < dist[que[0]]*num:
                        que.append(que.popleft())
                else:
                    su += r - dist[w]
                    dist[w] = r
    return dist
dist = SPFA(N, G, r)
if dist is None:
    print("NEGATIVE CYCLE")
else:
    for i in range(N):
        print(dist[i] if dist[i] < INF else "INF")
