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 MAX_V 1000
#define MAX_E 2000
#define INF 1e9

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

int V,E,d[MAX_V];
edge es[MAX_E];

bool find_negative_loop(int s){
  int cnt = 0;
  for(int i = 0 ; i < V ; i++){
    d[i] = INF;
  }
  d[s] = 0;
  while(true){
    bool update = false;
    cnt++;
    for(int i = 0 ; i < E ; i++){
      edge e = es[i];
      if(d[e.from] != INF && d[e.to] > d[e.from] + e.cost){
        d[e.to] = d[e.from] + e.cost;
        if(cnt == V){ return true; }
        update = true;
      }
    }
    if(!update){ break; }
  }
  return false;
}

int main(){
  int r;
  cin >> V >> E >> r;
  for(int i = 0 ; i < E ; i++){
    cin >> es[i].from >> es[i].to >> es[i].cost;
  }
  if(find_negative_loop(r)){
    cout << "NEGATIVE CYCLE" << endl;
  }else{
    for(int i = 0 ; i < V ; i++){
      if(d[i] == INF){
        cout << "INF" << endl;
      }else{
        cout << d[i] << endl;
      }
    }
  }
  return 0;
}