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 rep(i,n) for(int i=0;i<n;i++)
#define ll long long int
#define INF 1000000007
// cin.eof() 入力個数の問題
int main(){
int v, e, r;
    cin >> v >> e >> r;
    vector<pair<pair<int,int>,int>> G;
    rep(i, e) {
        int a, b, c;
        cin >> a >> b >> c;
        G.push_back(make_pair(make_pair(a,b),c));
    }
    vector<int> ans;
    rep(i, v) {
        ans.push_back(INF);
    }
    ans[r]=0;

    int count=0;
    rep(i,v){
      rep(k,e){
        int u=G[k].first.first,V=G[k].first.second,cost=G[k].second;
        //cout<<"u=="<<u<<" v=="<<V<<" cost=="<<cost<<" i=="<<i<<endl;
        //cout<<"ans[u]+cost=="<<ans[u]+cost<<" ans[V]=="<<ans[V]<<endl;
        if(ans[u]+cost<ans[V]&&ans[u]!=INF){
          //cout<<"change"<<endl;
          ans[V]=ans[u]+cost;
          if(i==v-1){
            cout<<"NEGATIVE CYCLE"<<endl;
            return 0;
          }
        }
      }
    }
    
rep(i,v){
  if(ans[i]==INF){
    cout<<"INF"<<endl;
  }else
  {
    cout<<ans[i]<<endl;
  }
  
}

  return 0;
}

