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
# -*- coding: utf-8 -*-
"""
Created on Sun Jun 30 11:59:44 2019
Bellman Ford Method
@author: maezawa
"""

import collections


class BellmanFord():
    def __init__(self):
        self.edges = []

    def add(self, u, v, d, directed=False):
        """
        u = from, v = to, d = cost
        directed = Trueのとき、有向グラフである。
        """
        if directed:
            self.edges.append([u, v, d])
        else:
            self.edges.append([u, v, d])
            self.edges.append([v, u, d])

    def BellmanFord_search(self, s):
        """
        次のBellmanFord_negative_boolまで考慮した時には、0-indexedであることに注意
        s =  始点
        return: d[i] 始点sから各点iまでの最短経路
        """
        d = collections.defaultdict(lambda: float('inf'))
        d[s] = 0
        numEdges = len(self.edges)
        while True:
            update = False
            for i in range(numEdges):
                e = self.edges[i]
                # e: 辺iについて [from,to,cost]
                if d[e[0]] != float("inf") and d[e[1]] > d[e[0]] + e[2]:
                    d[e[1]] = d[e[0]] + e[2]
                    update = True
            if not update:
                break
        return d

    def BellmanFord_negative_bool(self, numNodes, start, zero_indexed=False):
        # 負の閉路の検出, Trueなら負の閉路が存在する
        d = collections.defaultdict(lambda: float('inf'))
        d[start] = 0
        numEdges = len(self.edges)
        for i in range(numNodes):
            for j in range(numEdges):
                e = self.edges[j]
                if d[e[1]] > d[e[0]] + e[2]:
                    d[e[1]] = d[e[0]] + e[2]
                    #print(i, j, e)
                    if i == numNodes-1:
                        return True
        #print(d)
        return False
    

graph = BellmanFord()
num_v, num_edges, start = list(map(int, input().split()))

for i in range(num_edges):
    u, v, w = list(map(int, input().split()))
    graph.add(u, v, w, directed=True)

if graph.BellmanFord_negative_bool(num_v, start, zero_indexed=True):
    print('NEGATIVE CYCLE')
else:
    distances = graph.BellmanFord_search(start)
    for i in range(num_v):
        if distances[i] == float('inf'):
            print('INF')
        else:
            print(distances[i])

