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<iostream>
#include<vector>
#include<utility>
#include<algorithm>
using namespace std;

int main(){
  int V,E,r;
  cin >> V >> E >> r;
  vector< vector< pair<int,int> > > v(V);
  int s,t,d;
  for(int i = 0; i < E; ++i){
    cin >> s >> t >> d;
    v[t].push_back(make_pair(s,d));
  }
  vector<int> M(V,20000010);
  M[r] = 0;
  int i;
  for(i = 0; i < V; ++i){
    int check = 0;
    for(int j = 0; j < V; ++j){
      int minimum = 20000010;
      for(size_t k = 0; k < v[j].size(); ++k){
	int another;
	if(M[v[j][k].first] == 20000010) another = 20000010;
	else another = M[v[j][k].first] + v[j][k].second;
	minimum = min(minimum, another);
      }
      minimum = min(minimum, M[j]);
      if(M[j] == minimum) check += 1;
      M[j] = minimum;
    }
    if(check == V) break;
  }
  if(i == V) cout << "NEGATIVE CYCLE" << endl;
  else{
    for(int j = 0; j < V; ++j){
      if(M[j] == 20000010) cout << "INF" <<endl;
      else cout << M[j] << endl;
    }
  }
  return 0;
}