human
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
Solve the task in PYTHON3.
 Write me a Python code to print the cost of the shortest path from the source r to each vertex 0, 1,?
assistant
#Here is Python code to print the cost of the shortest path from the source r to each vertex 0, 1, .

class BellmanFord:
    INF = int(1e18)
    N = -1
    edge = []
    
    def __init__(self, N):
        self.N = N
        
    def add_edge(self, frm, to, cost):
        self.edge.append((frm, to, cost));
        
    def add_bi_edge(self, a, b, cost):
        self.add_edge(a, b, cost)
        self.add_edge(b, a, cost)
    
    def calcD(self, start):
        D = [self.INF] * self.N
        D[start] = 0
        update, cnt = True, 0
        while update:
            update = False
            cnt = cnt + 1
            if cnt > self.N:
                return 'NegativeCycle'
            
            for frm, to, cost in self.edge:
                ncost = D[frm] + cost
                if D[frm] == self.INF or ncost >= D[to]:
                    continue
                D[to] = ncost
                update = True
                
        return D

    
V, E, S = list(map(int, input().split()))
bf = BellmanFord(V)
for i in range(E):
    frm, to, cost = list(map(int, input().split()))
    bf.add_edge(frm, to, cost)
    

ans = bf.calcD(S)
if ans == 'NegativeCycle':
    print("NEGATIVE CYCLE")
else:
    for i in range(V):
        print ('INF' if ans[i] == bf.INF else ans[i])