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:
"""


Verified:
    https://onlinejudge.u-aizu.ac.jp/problems/GRL_1_B
"""
from typing import List, Optional, Tuple


def bellman_ford(
    source: int, vertices_count: int, edges: List[Tuple]
) -> Optional[List[float]]:
    """Bellman-Ford algorithm: O(NM)
    Compute the distance of the shortest path from V_source to each vertex.
    Relax edges in N-1 times of loops and detect negative cycles in the last loop.

    Args:
        source: V_source
        edges: Edges info
        vertices_count: Number of vertices
    Returns:
        Return None if the graph has negative cycles.
        If not, return a list of distance from V_source.
    """
    inf = float("inf")
    dist = [inf] * (vertices_count + 1)
    dist[source] = 0
    for i in range(vertices_count):
        for from_, to, weight in edges:
            if dist[from_] + weight < dist[to]:
                dist[to] = dist[from_] + weight
                if i == vertices_count - 1:
                    return None
    return dist


def main():
    V, E, R, *edges = map(int, open(0).read().split())
    dist = bellman_ford(R, V, list(zip(*[iter(edges)] * 3)))
    inf = float("inf")
    if dist:
        print("\n".join(map(str, (i if i != inf else "INF" for i in dist[:-1]))))
    else:
        print("NEGATIVE CYCLE")


if __name__ == "__main__":
    main()