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:
V, E, r = map(int, input().split())

INF = 2 << 21
G = [[] for i in range(V)]
dist = [INF for j in range(V)]


def bellman_ford(s):
    p, flg = [-1 for i in range(V)], [False for i in range(V)]
    dist[s], flg[s], cnt = 0, True, 0

    while True:
        if cnt >= 2 * V:
            return False

        on = []
        for i in range(V):
            if flg[i]:
                on.append(i)
                flg[i] = False

        if len(on) == 0:
            return True

        for i in on:
            for j in G[i]:
                if dist[i] + j[1] < dist[j[0]]:
                    dist[j[0]] = dist[i] + j[1]
                    p[j[0]], flg[j[0]] = i, True

        cnt += 1


def main():
    for i in range(E):
        s, t, d = map(int, input().split())
        G[s].append((t, d))

    check = bellman_ford(r)

    if check:
        for i in range(V):
            if dist[i] == INF:
                print("INF")
            else:
                print(dist[i])
    else:
        print("NEGATIVE CYCLE")


if __name__ == '__main__':
    main()