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 <vector>
using namespace std;
const int INF = 1001001001;
int V, E,r;
struct edge{int u,v,cost;};
vector<edge> A;
vector<int> d;

int main() {
    cin >> V >> E >> r;
    A.resize(E);
    d.resize(V);
    for(int i=0; i<E; i++) {
        int s,t,w;
        cin >> s >> t >> w;
        struct edge temp = {s,t,w};
        A[i] = temp;
    }

    for(int i=0; i<V; i++) d[i] = INF;
    d[r] = 0;

    for(int i=0; i<V-1; i++) {
        for(int j=0; j<E; j++) {
            struct edge e = A[j];
            if(d[e.u] == INF) continue;
            d[e.v] = min(d[e.v], d[e.u] + e.cost);
        }
    }

    for(int i=0; i<E; i++) {
        struct edge e = A[i];
        if(d[e.u] != INF && d[e.v] > d[e.u] + e.cost) {
            cout << "NEGATIVE CYCLE" << endl;
            return 0;
        }
    }

    for(int i=0; i<V; i++) {
        if(d[i] == INF) cout << "INF" << endl;
        else cout << d[i] << endl;
    }
    return 0;
}
