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<bits/stdc++.h>
using namespace std;
#define int long long 

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

int N,M;

const int INF = 1LL << 60;

int d[2020];

edge E[2020];

void Bellmanford(const int &a){
  
  for(int i = 0; i < N; i++){
    d[i] = INF;
  }
  
  d[a] = 0;
  
  for(int k = 0; k < N; k++){
    
    bool update = false;
    
    for(int i = 0; i < M; i++){ 
    
    if( d[E[i].from] != INF && d[E[i].to] > d[E[i].from]+E[i].cost ){
      d[E[i].to] = d[E[i].from]+E[i].cost;
      update  = true;
    }
  }
  
    if(update && k == N-1){ 
      cout << "NEGATIVE CYCLE" << endl; 
      exit(0);
    }
    
  }
  
  for(int i = 0; i < N; i++){ 
    int res = d[i];
    if(res == INF){ cout << "INF" << endl;}
    else{ cout << res << endl;}
  }
 
  return ;
}
    
    

signed main(void){
  
  int r; 
  
  cin >> N >> M >> r;
  
  for(int i = 0; i < M; i++){ 
    
    cin >> E[i].from >> E[i].to >> E[i].cost;
    
  }
  
  
  Bellmanford(r);
  
  return 0;
}
  
