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:
//ベルマン・フォード法
#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<queue>

using namespace std;

#define INF 1e9
using vvi = vector< vector<int> >;

int main(){
  int v, e, r;
  cin >> v >> e >> r;
  vector< pair<int, pair<int, int> > > edge;
  for(int i = 0; i < e; i++){
    int s,t,d;
    cin >> s >> t >> d;
    edge.push_back({s, {t,d}});
  }
  vector<int> d(v, INF);
  d[r] = 0;
  for(int i = 0; i < v; i++){
    for(int j = 0; j < e ; j++){
      if(d[edge[j].first] != INF && d[edge[j].second.first] > d[edge[j].first] + edge[j].second.second){
        d[edge[j].second.first] = d[edge[j].first] + edge[j].second.second;
        if(i == v-1){
          cout << "NEGATIVE CYCLE" << endl;
          return 0;
        }
      }
    }
  }
  for(int i = 0; i < v; i++){
    if(i == r){
      cout << 0 << endl;
    }
    else{
      if(d[i] == INF) cout << "INF" << endl;
      else cout << d[i] << endl;
    }
  }
}