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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
struct edge{int from, to; ll cost;};
const ll INF = 1e17;

bool bellman_ford(int n, int s, vector<edge>& es, vector<ll>& d){
  rep(i,n) d[i] = INF;
  d[s] = 0;
  bool negative_loop = false;
  rep(i,n) {
    bool update = false;
    for(auto e : es) {
      if(d[e.from] != INF && d[e.to] > d[e.from] + e.cost){
        if(i == n-1) negative_loop = true;
        d[e.to] = d[e.from] + e.cost;
        update = true;
      }
    }
    if(!update) break;
  }
  return negative_loop;
}

int main(){
  int v, e, r;
  cin >> v >> e >> r;
  vector<edge> es;
  rep(i,e) {
    int s, t; ll dist;
    cin >> s >> t >> dist;
    es.push_back(edge{s,t,dist});
  }
  vector<ll> d(v);
  bool f = bellman_ford(v,r,es,d);
  if(f) cout << "NEGATIVE CYCLE" << endl;
  else rep(i,v) cout << (d[i] == INF ? "INF" : to_string(d[i])) << endl;
  return 0;
}