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<map>
#include<string>
#include<vector>
#include<algorithm>
#include<numeric>
#include<cmath>
#include<queue>
using namespace std;
int inf = 1000000000;

int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};

struct Node{
  int to, from, cost;
};

int main(){
  int n, m, r;
  cin >> n >> m >> r;
  vector<int> dist(n, inf);
  vector<Node> node(m);
  for(int i=0; i<m; i++){
    int s, t, d;
    cin >> s >> t >> d;
    node[i].to = t; node[i].from = s; node[i].cost = d;
  }

  dist[r] = 0;
  int l = 0;
  while( true ){
    if( l > 100000 ){
      cout << "NEGATIVE CYCLE" << endl;
      return 0;
    }
    bool f = false;
    for(int i=0; i<node.size(); i++){
      Node d = node[i];
      if( dist[d.from] != inf && dist[d.to] > dist[d.from] + d.cost ){
        dist[d.to] = dist[d.from] + d.cost;
        f = true;
      }
    }
    if( !f ) break;
    l++;
  }

  for(int i=0; i<n; i++){
    if( dist[i] == inf ) cout << "INF" << endl;
    else cout << dist[i] << endl;
  }

  return 0;
}





// EOF