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 <vector>
using namespace std;


struct Edge {
    int to, cost;

};
typedef vector<vector<Edge> > AdjList;  // 隣接リストの型
AdjList graph;

const int INF = 1000000;

vector<int> dist;


bool bellman_ford(int n, int s){
    dist = vector<int>(n,INF);
    dist[s] = 0;
    for(int i=0;i<n;i++){
        for(int v=0;v<n;v++){
            for(int k=0;k<graph[v].size();k++){
                Edge e = graph[v][k];
                if(dist[v] != INF && dist[e.to] > dist[v] + e.cost){
                    dist[e.to] = dist[v] + e.cost;
                    if(i == n-1) return true;
                }
            }
        }
    }
    return false;
}

int main(void){
    int n, m, r;
    cin >> n >> m >> r;

    graph = AdjList(n);

    for(int i=0;i<m;i++){
        int from, to, cost;
        cin >> from >> to >> cost;
        graph[from].push_back((Edge){to,cost});
    }
    bool negative;
    negative = bellman_ford(n,r);
    if(negative){
      cout << "NEGATIVE CYCLE" << endl;
      return 0;
    }
    for (int i = 0; i < n; i++){
       if(dist[i] == INF) cout << "INF" << endl;
       else cout << dist[i] << endl;
    }
    return 0;
}