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;
static const int MAX_V = 1000;
static const int MAX_E = 2000;
static const int INF = 1 << 30;

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

edge es[MAX_E];

int d[MAX_V];
int V, E;
bool def = true;

void shortest_path(int s){
	for(int i = 0; i < V; i++) d[i] = INF;
	d[s] = 0;
	int cnt = 0;
	while(true){
		cnt++;
		bool update = false;
		for(int i = 0; i < E; i++){
			edge e = es[i];
			if(d[e.from] != INF && d[e.to] > d[e.from] + e.cost){
				d[e.to] = d[e.from] + e.cost;
				update = true;
			}
		}
		if(!update) break;
		if(cnt == V){
			def = false;
			return;
		}
	}
}

int main(){
	int r;
	scanf("%d %d %d", &V, &E, &r);
	for(int i = 0; i < E; i++){
		scanf("%d %d %d", &es[i].from, &es[i].to, &es[i].cost);
	}
	shortest_path(r);
	if(!def) printf("NEGATIVE CYCLE\n");
	else{
		for(int i = 0; i < V; i++){
			if(d[i] == INF) printf("INF\n");
			else printf("%d\n", d[i]);
		}
	}
	return 0;
}