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 <iostream>
#include <vector>
#define MAX_E 2000
#define MAX_V 1000
#define INF 1000000
using namespace std;

typedef struct {
  int from, to, cost;
} edge;

edge es[MAX_E];

int d[MAX_V];

bool bellman_ford(int v, int e, int start) {
  d[start] = 0;
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < e; j++) {
      edge temp = es[j];
      if (d[temp.from] == INF) continue;
      if (d[temp.to] > d[temp.from] + temp.cost) {
	d[temp.to] = d[temp.from] + temp.cost;
	if (i == v - 1) return true;
      }
    }
  }
  return false;
}

void init() {
  for (int i = 0; i < MAX_V; i++) {
    d[i] = INF;
  }
}

int main() {
  int v, e, r;
  cin >> v >> e >> r;
  for (int i = 0; i < e; i++) {
    int s, t, d;
    cin >> s >> t >> d;
    edge temp;
    temp.from = s;
    temp.to = t;
    temp.cost = d;
    es[i] = temp;
  }

  init();

  if (bellman_ford(v, e, r) == true) {
    cout << "NEGATIVE CYCLE" << endl;
    return 0;
  }
  for (int i = 0; i < v; i++) {
    if (d[i] == INF) cout << "INF" << endl;
    else cout << d[i] << endl;
  }
  return 0;
}