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 <bits/stdc++.h>
using namespace std;

const int maxv = 1005, maxe = 2005;

int md[maxv], pt[maxv];
int q[maxv], qh = 0, qt = 0;
bool inq[maxv];
vector<pair<int, int>> G[maxv];

bool empty() { return qh == qt; }

void push(int v) {
	if (inq[v]) return;
	inq[v] = true;
	q[qt] = v;
	qt++;
	if (qt >= maxv) qt -= maxv;
}

int pop() {
	int v = q[qh++];
	inq[v] = false;
	if (qh >= maxv) qh -= maxv;
	return v;
}

int main() {
	int v, e, r; cin >> v >> e >> r;
	
	for (int i = 0; i < e; i++) {
		int s, t, d; cin >> s >> t >> d;
		G[s].push_back({t, d});
		//G[t].push_back({s, d});
	}
	
	bool ok = true;
	fill(md, md + maxv, 0x3f3f3f3f);
	
	md[r] = 0;
	push(r);
	
	while (!empty()) {
		int t = pop();
		pt[t]++;
		if (pt[t] > maxe) {
			ok = false;
			break;
		}
		for (pair<int, int> p : G[t]) {
			if (md[t] + p.second < md[p.first]) {
				md[p.first] = md[t] + p.second;
				push(p.first);
			}
		}
	}
	
	if (!ok) {
		cout << "NEGATIVE CYCLE" << endl;
	} else {
		for (int i = 0; i < v; i++) {
			if (md[i] >= 0x3f3f3f3f) {
				cout << "INF" << endl;
			} else {
				cout << md[i] << endl;
			}
		}
	}
}