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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;

const int INF = 1001001001;

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);

  int v, e, r;
  cin >> v >> e >> r;

  vector<int> s(e, INF), t(e, INF), w(e, INF);
  for (int i = 0; i < e; i++) {
    cin >> s[i] >> t[i] >> w[i];
  }

  vector<vector<P>> adj(e);
  for (int i = 0; i < e; i++) {
    adj[s[i]].push_back(P(t[i], w[i]));
  }

  vector<int> d(v, INF);
  d[r] = 0;

  int cnt = 0;
  while (cnt < v) {
    bool updated = false;
    for (int i = 0; i < e; i++) {
      int from = i;
      for (int j = 0; j < adj[i].size(); j++) {
        int to = adj[i][j].first;
        int cost = adj[i][j].second;

        if (d[from] == INF) continue;

        if (d[from] + cost < d[to]) {
          d[to] = d[from] + cost;
          updated = true;
        }
      }
    }

    if (!updated) {
      break;
    }

    cnt++;
  }

  if (cnt == v) {
    cout << "NEGATIVE CYCLE" << endl;
  } else {
    for (int i = 0; i < v; i++) {
      if (d[i] < INF) {
        cout << d[i] << endl;
      } else {
        cout << "INF" << endl;
      }
    }
  }

  return 0;
}
