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<cstdio>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<ll, P> Q;

ll v, e, r, s[2000], t[2000], d[2000], dp[1001], INF = 1e9;

bool search() {
  for (ll i = 0; i < v; i++) dp[i] = INF;
  dp[r] = 0;
  for (ll i = 0; i < v; i++) {
    bool ok = true;
    for (ll j = 0; j < e; j++) {
      if (dp[s[j]] == INF) continue;
      if (dp[t[j]] > dp[s[j]] + d[j]) {
        dp[t[j]] = dp[s[j]] + d[j];
        ok = false;
        if (i == v-1) return true;
      }
    }
    if (ok) break;
  }
  return false;
}

int main() {
  scanf("%lld%lld%lld", &v, &e, &r);
  for (ll i = 0; i < e; i++) scanf("%lld%lld%lld", &s[i], &t[i], &d[i]);
  if (search()) {
    printf("NEGATIVE CYCLE\n");
  } else {
    for (ll i = 0; i < v; i++) {
      if (dp[i] == INF) printf("INF\n");
      else printf("%lld\n", dp[i]);
    }
  }
}