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:
#include <iostream>
#include <string>
#include <queue>
#include <vector>
#include <algorithm>
#include <utility>

#define INF 1000000000000000

using namespace std;

typedef long long int lli;
typedef pair<int, int> P;

struct edge{
	int from, to, cost;
};

int V, E, r;
edge e[2000];
lli d[1000];

bool solve(int s){
	fill(d, d+V, INF);
	d[s] = 0;
	for(int i = 0;i < V;i++){
		bool update = false;
		for(int j = 0;j < E;j++){
			if(d[e[j].from] != INF && d[e[j].to] > d[e[j].from] + e[j].cost){
				d[e[j].to] = d[e[j].from] + e[j].cost;
				update = true;
			}
		}
		if(!update) break;
		if(i == V-1 && update){
			return false;
		}
	}
	return true;
}

int main(void){
	cin >> V >> E >> r;
	for(int i = 0;i < E;i++){
		cin >> e[i].from >> e[i].to >> e[i].cost;
	}
	if(solve(r)){
		for(int i = 0;i < V;i++){
			if(d[i] != INF)
				cout << d[i] << endl;
			else
				cout << "INF" << endl;
		}
	}else{
		cout << "NEGATIVE CYCLE" << endl;
	}
	return 0;
}