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 <stdio.h>
#include <vector>
#include <utility>
typedef long long ll;

int main(void) {
  ll inf = 10e15;
  ll i, j, k, v, r, e;
  scanf("%lld%lld%lld", &v, &e, &r);
  ll dis[v];
  std::vector< std::pair<ll, ll> > vec[v];
  std::vector< std::pair<ll, ll> >::iterator it;
  for(i = 0; i < v; ++i) dis[i] = inf;
  dis[r] = 0;
  for(i = 0; i < e; ++i) {
    ll ta, tb, tc;
    std::pair<ll, ll> p;
    scanf("%lld%lld%lld", &ta, &tb, &tc);
    p = std::make_pair(tb, tc);
    vec[ta].push_back(p);
  }
  for(k = 1; k < v; ++k) {
    char flag = 0;
    for(i = 0; i < v; ++i) if(dis[i] != inf) {
      for(it = vec[i].begin(); it != vec[i].end(); ++it) {
        if(dis[(*it).first] > dis[i] + (*it).second) {
          flag = 1;
          dis[(*it).first] = dis[i] + (*it).second;
        }
      }
    }
    if(flag) {
      if(k == v - 1) {
        printf("NEGATIVE CYCLE\n");
        return 0;
      }
    } else break;
  }
  for(i = 0; i < v; ++i) {
    if(dis[i] != inf) printf("%lld\n", dis[i]);
    else printf("INF\n");
  }
  return 0;
}