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;
const int INF = 1 << 28;
typedef pair<int, int> Pr;

int n, m, start;
vector<Pr> g[1000];
int dist[1000];

int solve()
{
	for (int i = 0; i < n; i++){
		dist[i] = INF;
	}
	dist[start] = 0;
	
	int k = 0;
	while (k < n){
		bool update = false;
		for (int i = 0; i < n; i++){
			if (dist[i] == INF) continue;
			for (int j = 0; j < g[i].size(); j++){
				int from = i;
				int to = g[i][j].first;
				int cost = g[i][j].second;
				if (dist[to] > dist[from] + cost){
					dist[to] = dist[from] + cost;
					update = true;
				}
			}
		}
		if (!update) break;
		k++;
	}
	
	if (k == n) return -1;
	return 1;
}

int main()
{
	scanf("%d %d %d", &n, &m, &start);
	
	for (int i = 0; i < m; i++){
		int s, t, d;
		scanf("%d %d %d", &s, &t, &d);
		
		g[s].push_back(Pr(t, d));
	}
	
	if (solve() == -1){
		puts("NEGATIVE CYCLE");
	}
	else {
		for (int i = 0; i < n; i++){
			if (dist[i] == INF) puts("INF");
			else printf("%d\n", dist[i]);
		}
	}
}