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;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define inf 1000000000000000
#define mod 1000000007
typedef pair<int, int> P;

bool prime(int n) {
	for (int i = 2; i <= sqrt(n); i++) {
		if (n % i == 0)return false;
	}
	return n != 1;
}
int gcd(int x, int y) {
	if (y == 0)return x;
	return gcd(y, x % y);
}
int lcm(int x, int y) {
	return x / gcd(x, y) * y;
}

struct edge {
	int from, to, cost;
};

edge g[3000];
int d[1050];
int V, E, S;

signed main() {
	cin >> V >> E >> S;
	rep(i, E) {
		cin >> g[i].from >> g[i].to >> g[i].cost;
	}
	fill(d, d + V, inf);
	d[S] = 0;
	for (int i = 0; i < V; i++) {
		bool update = false;
		rep(j, E) {
			edge e = g[j];
			if (d[e.from] != inf && d[e.from] + e.cost < d[e.to]) {
				d[e.to] = d[e.from] + e.cost;
				update = true;
			}
		}
		if (!update) {
			break;
		}
		if (i == V - 1) {
			cout << "NEGATIVE CYCLE" << endl;
			return 0;
		}
	}
	rep(i, V) {
		if (d[i] == inf) {
			cout << "INF" << endl;
		}
		else {
			cout << d[i] << endl;
		}
	}

	return 0;
}

