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;

bool bellman_ford(const int v, const int s,
                  const vector<pair<pair<int, int>, int64_t>>& g,
                  vector<int64_t>& cost)
{
    const int64_t inf = 1LL << 60;
    cost[s] = 0;
    for (auto i = 0; i < v; i++) {
        for (auto & elm : g) {
            if(cost[elm.first.first] == inf) continue;
            if(cost[elm.first.second] > cost[elm.first.first] + elm.second) {
                if(i == v - 1) return true; // 負の閉路がある
                cost[elm.first.second] = cost[elm.first.first] + elm.second;
            }
        }
    }
    return false;
}

int main(){
    int v, e, r; cin >> v >> e >> r;
    vector<pair<pair<int, int>, int64_t>> g;
    const int64_t inf = 1LL << 60;
    vector<int64_t> cost(v, inf);
    for (auto i = 0; i < e; i++) {
        int s, t, d; cin >> s >> t >> d;
        g.emplace_back(make_pair(s, t), d);
    }
    if(bellman_ford(v, r, g, cost)) {
        cout << "NEGATIVE CYCLE" << endl;
        return 0;
    }
    for (auto & elm : cost)
        cout << (elm == inf ? "INF" : to_string(elm)) << endl;
    return 0;
}