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<cstdio>
#include<vector>
using namespace std;

struct edge{int from,to,cost;};
const int INF = 1000000000;
int v,e,r;
vector<edge> E;
int d[1000];

int main(){
  scanf("%d %d %d",&v,&e,&r);
  for(int i=0;i<e;i++){
    edge tmp;
    scanf("%d %d %d",&tmp.from,&tmp.to,&tmp.cost);
    E.push_back(tmp);
  }
  fill(d,d+v,INF);
  d[r] = 0;
  int t = 0;
  while(true){
    bool flag = true;
    for(int i=0;i<E.size();i++){
      if(d[E[i].from]!=INF&&d[E[i].from]+E[i].cost<d[E[i].to]){
        d[E[i].to] = d[E[i].from]+E[i].cost;
        flag = false;
      }
    }
    if(flag) break;
    if(t>v+5){
      printf("NEGATIVE CYCLE\n");
      return 0;
    }
    t++;
  }
  for(int i=0;i<v;i++){
    if(d[i]==INF) printf("INF\n");
    else printf("%d\n",d[i]);
  }
  return 0;
}