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 <iostream>
#include <queue>
using namespace std;

int n, m, s;
vector<int> et[1000];
vector<int> ec[1000];
int minDist[1000];
int INF = 1e+9;

int main() {
	int i;
	
	cin >> n >> m >> s;
	for (i = 0; i < m; i++) {
		int u, v, d;
		cin >> u >> v >> d;
		et[u].push_back(v);
		ec[u].push_back(d);
	}
	
	for (i = 0; i < n; i++) minDist[i] = INF;
	
	queue<int> que;
	
	minDist[s] = 0;
	que.push(s);
	
	int loopCnt = 0;
	while (!que.empty()) {
		if (loopCnt > n * m + 100) {
			cout << "NEGATIVE CYCLE" << endl;
			return 0;
		}
		int v = que.front(); que.pop();
		for (i = 0; i < et[v].size(); i++) {
			int nv = et[v][i];
			if (minDist[nv] > minDist[v] + ec[v][i]) {
				minDist[nv] = minDist[v] + ec[v][i];
				que.push(nv);
			}
		}
		loopCnt++;
	}
	
	for (i = 0; i < n; i++) {
		if (minDist[i] >= INF) {
			cout << "INF" << endl;
		}
		else {
			cout << minDist[i] << endl;
		}
	}
	return 0;
}
