Problem: 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

Solution (in CPP):

Solution:
// GRL_1_B
#include <iostream>
#include <vector>
#include <queue>
#include <math.h>
#define   N   1000000000
using namespace std;

struct edge{
	int adj;
	int dis;
};

int main(){
	int V, E, r, s, t, d, x, count=0;
	vector<vector<edge> > A;
	vector<int> distance;
	queue<int> ver;

	edge e;

	cin>>V>>E>>r;

	A.resize(V);
	for(int i=0;i<V;i++){
		distance.push_back(N);
	}
	distance[r] = 0;

	for(int j=0;j<E;j++){
		cin>>s>>t>>d;
		e.adj = t;
		e.dis = d;
		A[s].push_back(e);
	}
	
	ver.push(r);

	while(count<=V){
		queue<int> vern;
		while(!ver.empty()){
			x = ver.front();
			for(int i=0;i<A[x].size();i++){
				if(distance[A[x][i].adj] > distance[x] + A[x][i].dis){
					distance[A[x][i].adj] = distance[x] + A[x][i].dis;
					vern.push(A[x][i].adj);
				}
			}
			ver.pop();
		}
		if(vern.empty()) break;
		ver = vern;
		count++;
	}

	if(count > V){
		cout<<"NEGATIVE CYCLE"<<endl;
	}else{
		for(int i=0;i<V;i++){
			if(distance[i] >= N)
				cout<<"INF"<<endl;
			else
				cout<<distance[i]<<endl;
		}
	}
	return 0;
}