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 = 1010;
const int MAXE = 2010;
const int64_t INF = (1LL << 62);
struct edge
{
  int from, to, cost;
};
edge es[MAXE];

vector<int64_t> d(MAXV, INF);
bool BellmanFord(int s, int V, int E)
{
  d[s] = 0;
  for (int i = 0; i < V; i++)
  {
    for (int j = 0; j < E; j++)
    {
      edge e = es[j];
      if (d[e.from] == INF)
        continue;
      if (d[e.to] > d[e.from] + e.cost)
      {
        d[e.to] = d[e.from] + e.cost;
        if (i == V - 1)
        {
          return false;
        }
      }
    }
  }
  return true;
}

int main()
{
  int V, E, r, v, u, c;
  cin >> V >> E >> r;
  for (int i = 0; i < E; i++)
  {
    cin >> v >> u >> c;
    es[i] = {v, u, c};
  }
  if (BellmanFord(r, V, E))
  {
    for (int i = 0; i < V; i++)
    {
      if (d[i] == INF)
        cout << "INF" << endl;
      else
        cout << d[i] << endl;
    }
  }
  else
  {
    cout << "NEGATIVE CYCLE" << endl;
  }
}