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>
using namespace std;
const int INFTY=1<<28;
const int MAX=1000;

int n,m;//number of vertex,edge
int d[MAX];
vector<pair<int,int>> V[MAX];//to,cost

bool bellman_ford(int s)//false is negative cycle
{
	for(int i=0;i<n;i++)d[i]=INFTY;
	d[s]=0;
	bool update=true;
	pair<int,int>p;
	int c=0;
	
	for(int k=0;update&&k<m;k++)
	{
		update=false;
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<V[i].size();j++)
			{
				p=V[i][j];
				if(d[i]!=INFTY&&d[p.first]>d[i]+p.second)
				{
					d[p.first]=d[i]+p.second;
					update=true;
					if(k==m-1)return false;
				}
			}
		}
	}
	
	return true;
}

int main() {
	cin>>n;
	int r;cin>>m>>r;
	int a,b,c;
	for(int i=0;i<m;i++)
	{
		cin>>a>>b>>c;
		V[a].push_back(make_pair(b,c));
	}
	bool flag=bellman_ford(r);
	if(flag)
	{
		for(int i=0;i<n;i++)
		{
			if(d[i]!=INFTY)cout<<d[i]<<endl;
			else cout<<"INF"<<endl;
		}
	}
	else cout<<"NEGATIVE CYCLE"<<endl;
	return 0;
}