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 all(c) (c).begin(),(c).end()
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define iter(c) __typeof((c).begin())
#define tr(it,c) for(iter(c) it=(c).begin(); it!=(c).end(); it++)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define pr(a) cout << (a) << endl
typedef long long ll;
typedef pair<int,int> P;
const int MAX=1000000000;
int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};
struct edge{int from,to,cost;};
vector<edge> G;
int d[2001];

int main() {
  int n,m,s;
  cin >> n >> m >> s;
  rep(i,m) {
    edge e;
    cin >> e.from >> e.to >> e.cost;
    G.pb(e);
  }
  fill(d,d+n,MAX);
  d[s]=0;
  bool ck=false;
  for(int j=0; j<n; j++) {
    bool update=false;
    for(int i=0; i<m; i++) {
      edge e=G[i];
      if(d[e.from]!=MAX && d[e.to]>d[e.from]+e.cost) {
	d[e.to]=d[e.from]+e.cost;
	update=true;
	if(j==n-1) ck=true;
      }
    }
    if(!update) break;
  }
  if(ck) pr("NEGATIVE CYCLE");
  else {
    rep(i,n) {
      if(d[i]==MAX) pr("INF");
      else pr(d[i]);
    }
  }
  return 0;
}
