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;
typedef long long ll;
typedef pair<int,int> pii;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define all(a)  (a).begin(),(a).end()
#define pb push_back
#define INF (1e9+1)
//#define INF (1LL<<59)





#define MAX_V 1000

struct edge{int to,cost;};
vector<edge> G[MAX_V];

int shortest_path(int s,vector<int> &d){
	int v = d.size();
	rep(i,d.size())d[i]=INF;
	d[s]=0;
	rep(loop,v){
		bool update = false;
		rep(i,MAX_V){
			for(auto e:G[i]){
				if(d[i]!=INF && d[e.to] > d[i]+e.cost){
					d[e.to] = d[i]+e.cost;
					update = true;
				}
			}
		}
		if(!update)break;
		if(loop==v-1)return true;	//negative_cycle
	}
	return false;
}


int main(){
	int v,e,r;
	cin>>v>>e>>r;
	
	rep(i,e){
		int u,v,d;
		cin>>u>>v>>d;
		G[u].pb({v,d});
	}
	
	vector<int> d(v);
	bool res = shortest_path(r,d);
	if(res==true)cout<<"NEGATIVE CYCLE"<<endl;
	else{
		rep(i,v){
			if(d[i]==INF)cout<<"INF"<<endl;
			else cout<<d[i]<<endl;
		}
	}
}