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
#include <bits/stdc++.h>
using namespace std;

const int INF = 1e9;
struct edge{
    int to, cost;
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int v, e, r;
    cin >> v >> e >> r;

    vector<edge> G[v];
    for(int i = 0; i < e; i++){
        int s, t, d;
        cin >> s >> t >> d;
        G[s].push_back({t, d});
        //G[t].push_back({s, d}); いらんのです????????
    }

    int D[v];
    for(int i = 0; i < v; i++){
        D[i] = INF;
    }
    D[r] = 0;

    bool f = true;
    int cnt = 0;
    while(f){
        if(cnt > v) break;
        cnt++;
        f = false;
        for(int i = 0; i < v; i++){
            for(int j = 0; j < G[i].size(); j++){
                edge e = G[i][j];
                if(D[e.to] > D[i] + e.cost and D[i] != INF){
                    D[e.to] = D[i] + e.cost;
                    f = true;
                }
            }
        }
    }

    if(cnt > v){
        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;
}
