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;

struct edge {
  edge () {}
  edge (int to, int cost) : to(to), cost(cost) {}

  int to, cost;
};

bool bellman_ford(int s, const vector<vector<edge>> &G, vector<long long> &d) {
  const long long linf = 1001001001001001001ll;
  d = vector<long long>(G.size(), linf);
  d[s] = 0;
  bool f = false;

  for (int i = 0; i < G.size(); ++i) {
    for (int v = 0; v < G.size(); ++v) {
      for (auto &e : G[v]) {
        if (d[v] != linf && d[e.to] > d[v] + e.cost) {
          d[e.to] = d[v] + e.cost;
          if (i == G.size() - 1) f = true;
        }
      }
    }
  }

  return f;
}

int main() {
  int v, e, r;
  cin >> v >> e >> r;

  vector<vector<edge>> G(v);
  for (int i = 0; i < e; ++i) {
    int a, b, c;
    cin >> a >> b >> c;

    G[a].push_back(edge(b, c));
  }

  vector<long long> d;
  if (bellman_ford(r, G, d)) {
    cout << "NEGATIVE CYCLE" << endl;
  } else {
    for (int i = 0; i < v; ++i) {
      if (d[i] == 1001001001001001001ll) cout << "INF" << endl;
      else cout << d[i] << endl;
    }
  }
}