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>
#define ll long long
#define INF 100000000000000000
#define MAX 100010
using namespace std;

struct edge{
  int to;
  ll cost;
};

typedef pair<ll, int> p;
vector<edge> graph[MAX];
vector<ll> dist;

bool bellman_ford(int n, int s){ //n???????????°???s???????§???????
  dist = vector<ll>(n, INF); //dist???INF??§?????????
  dist[s] = 0;
  for(int i = 0; i < n; i++){
    for(int v = 0; v < n; v++){
      for(int k = 0; k < graph[v].size(); k++){
        edge e = graph[v][k];
        if(dist[v] != INF && dist[e.to] > dist[v] + e.cost){
          dist[e.to] = dist[v] + e.cost;
          if(i == n - 1) return true; // ????????????????????¨
        }
      }
    }
  }
  return false;
}

int main(){
  int v, e, r, s, t;
  ll d;
  cin >> v >> e >> r;
  for(int i = 0; i < e; i++){
    cin >> s >> t >> d;
    graph[s].push_back((edge){t, d});
  }
  bool check = bellman_ford(v, r);
  if(check){
    cout << "NEGATIVE CYCLE" << endl;
  }else{
    for(int i = 0; i < v; i++){
      if(dist[i] == INF) cout << "INF" << endl;
      else cout << dist[i] << endl;
    }
  }
  return 0;
}
