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 long long INF = 123456789987654321;
const int MAX = 123456;
struct edge {
	long long cost;
	int from, to;
};
int v, e, r;
edge ed[MAX];
long long dist[MAX];
void bellmanford(int s) {
	fill(dist, dist + MAX, INF);
	dist[s] = 0;
	int cnt = 0;
	while(true) {
		bool up = false;
		cnt++;
		for(int i = 0; i < e; i++) {
			edge tmp = ed[i];
			if(dist[tmp.from] != INF && dist[tmp.to] > dist[tmp.from] + tmp.cost) {
				dist[tmp.to] = dist[tmp.from] + tmp.cost;
				up = true;
			}
		}
		if(!up) {
			break;
		}
		if(cnt == v) {
			cout << "NEGATIVE CYCLE" << endl;
			exit(0);
		}
	}
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);  
	cin >> v >> e >> r;
	for(int i = 0; i < e; i++) {
		int s, t, d;
		cin >> s >> t >> d;
		ed[i] = edge{d, s, t};
	}
	bellmanford(r);
	for(int i = 0; i < v; i++) {
		if(dist[i] == INF) {
			cout << "INF" << endl;
		}
		else {
			cout << dist[i] << endl;
		}
	}
	return 0;	
}

