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;
using ll = long long;

struct Edge { int to; ll cost; };
vector<Edge> G[1010];
ll INF = 1e18;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int V, E; cin >> V >> E;
  int r; cin >> r;
  vector<ll> d(V, INF);

  for (int i = 0; i < E; i++) {
    int a, b; cin >> a >> b;//  a--; b--;
    ll c; cin >> c;
    G[a].emplace_back(Edge({b, c}));
    // G[b].emplace_back(Edge({a, c}));
  }

  d[r] = 0;
  for (int t = 0; t < V * 2; t++) {
    for (int i = 0; i < V; i++) {
      if (d[i] == INF) continue;
      for (auto e: G[i]) {
        if (d[e.to] > d[i] + e.cost) {
          d[e.to] = d[i] + e.cost;
          if (t > V) {
            cout << "NEGATIVE CYCLE" << '\n';
            return 0;
          }
        }
      }
    }
  }
  for (ll i = 0; i < V; i++) {
    ll ans = d[i];
    if (ans == INF) cout << "INF" << '\n';
    else cout << ans << '\n';
  }

  return 0;
}
