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 <stdio.h>
#include <utility>
#include <vector>
#include <queue>
using namespace std;

typedef long long int ll;
constexpr int kN = int(1E5 + 10), kM = int(5E5 + 10);
constexpr ll kInf = ll(1E16 + 10);

#define F first
#define S second
#define PB push_back

struct Edge {
	int to, cost;
};

vector<Edge> graph[kN];
ll dis[kN];

int main() {
	int n, m, s, l, r, c, cnt = 0;
	pair<ll, int> nxt;
	priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> pq;
	scanf("%d%d%d", &n, &m, &s);
	for (int i = 0; i < m; i++) {
		scanf("%d%d%d", &l, &r, &c);
		graph[l].PB({r, c});
	}
	
	for (int i = 0; i < n; i++) dis[i] = kInf;
	dis[s] = 0;
	pq.push({0, s});
	while (!pq.empty()) {
	    if (cnt++ > m * n) {
	        printf("NEGATIVE CYCLE\n");
	        return 0;
	    }
		nxt = pq.top();
		pq.pop();
		if (nxt.F > dis[nxt.S]) continue;
		for (Edge i : graph[nxt.S]) if (dis[nxt.S] + i.cost < dis[i.to]) {
			dis[i.to] = dis[nxt.S] + i.cost;
			pq.push({dis[i.to], i.to});
		}
	}
	for (int i = 0; i < n; i++) if (dis[i] == kInf) printf("INF\n");
	else printf("%lld\n", dis[i]);
}

