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
input = sys.stdin.readline


def bellman_ford(start: int, graph: list):
    """bellman-ford法: 始点startから各頂点への最短距離を求める(負辺OK)
    計算量: O(|E||V|)
    """
    INF = float("inf")
    n = len(graph)
    dist = [INF] * n
    dist[start] = 0
    if n == 1:
        return False, [0]
    for _ in range(n - 1):
        update = False
        for v, edges in enumerate(graph):
            for nxt_v, cost in edges:
                if dist[nxt_v] > dist[v] + cost:
                    dist[nxt_v] = dist[v] + cost
                    update = True
        if not update:
            break
    else:
        return True, dist  # 負閉路含む
    return False, dist  # 負閉路含まない

v, e, r = map(int, input().split())
info = [list(map(int, input().split())) for i in range(e)]
INF = 10 ** 18

graph = [[] for i in range(v)]
for a, b, cost in info:
    graph[a].append((b, cost))

is_negative, dist = bellman_ford(r, graph)

if is_negative:
    print("NEGATIVE CYCLE")
else:
    for num in dist:
        if num > 10 ** 12:
            print("INF")
        else:
            print(num)