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
#include <queue>
#include <iostream>
#include <algorithm>
#define ll long long
#define INF 100000000000000
using namespace std;
int main(){
  ios_base::sync_with_stdio(false);
  ll V, E, r;
  cin >> V >> E >> r;
  vector< vector< pair<ll,ll> > > edges(V);
  ll dist[V];
  for(ll i = 0; i < V; i++){
    dist[i] = INF;
  }
  for(ll i = 0; i < E; i++){
    ll s,t,d;
    cin >> s >> t >> d;
    edges[s].push_back(make_pair(t,d));
  }

  dist[r] = 0;

  for(ll i = 0; i < V; i++){
    for(ll k = 0; k < V; k++){
      if(dist[k] != INF){
	for(pair<ll,ll> edge: edges[k]){
	  if(dist[edge.first] > dist[k] + edge.second){
	    if(i == V - 1){
	      cout << "NEGATIVE CYCLE" << endl;
	      return 0;
	    }
	    dist[edge.first] = dist[k] + edge.second;
	  }
	}
      }
    }
  }
  for(ll i = 0; i < V; i++){
    if(dist[i] == INF)
      cout << "INF" << endl;
    else
      cout << dist[i] << endl;
  }
}