Problem: 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

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<functional>
#include<utility>

using namespace std;
typedef pair<int, int> P;
const int INF = (1 << 30);
int V, E, r;
int dp[100000];
struct edge{
	int to, cost;
	edge(int to,int cost=1):to(to),cost(cost){}
};
vector<edge> G[100000];

int main()
{
	cin.tie(0);
	ios::ios_base::sync_with_stdio(false);
	cin >> V >> E >> r;
	int s, t, d, update = 1, limit = V*E + 1;
	for (int i = 0; i < E; ++i) {
		cin >> s >> t >> d;
		edge e(t,d);
		G[s].emplace_back(e);
	}
	fill(dp, dp + V, INF);
	dp[r] = 0;
	priority_queue<P, vector<P>, greater<P>> que;
	que.push(P(0, r));
	while (!que.empty()) {
		P p = que.top(); que.pop();
		if (dp[p.second] < p.first)continue;
		int v = p.second;
		for (int i = 0; i < G[v].size(); ++i) {
			edge e = G[v][i];
			if (dp[e.to]>dp[v] + e.cost) {
				dp[e.to] = dp[v] + e.cost;
				++update;
				que.push(P(dp[e.to], e.to));
			}
		}
		if (update > limit) {
			cout << "NEGATIVE CYCLE\n";
			return 0;
		}
	}
	for (int i = 0; i < V; ++i) {
		if (dp[i] == INF)cout << "INF\n";
		else cout << dp[i] << "\n";
	}
	cout << flush;
	return 0;
}