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;

const int INF=INT_MAX;

struct bellford
{
    struct edge{int u, v, w;};
    int nv, ne;
    vector<int> d;
    vector<edge> E;

    bellford(int nv, int ne) : nv(nv), ne(ne)
    {
        d.resize(nv, INF);
    }

    void relax(int u, int v, int w)
    {
        if(d[u]!=INF && d[v]>d[u]+w) d[v]=d[u]+w;
    }

    bool get(int s)
    {
        d[s]=0;
        for(int i=0; i<nv-1; i++)
            for(auto &e : E)
                relax(e.u, e.v, e.w);
        
        for(auto &e : E)
            if (d[e.u]!=INF && d[e.v]>d[e.u]+e.w)
                return false;
        return true;
    }
};

int main()
{
    int nv, ne, r;
    cin>> nv >> ne >> r;
    bellford g(nv, ne);
    for(int i=0; i<ne; i++)
    {
        int u, v, w;
        cin >> u >> v >> w;
        g.E.push_back({u, v, w});
    }
    if(!g.get(r)) printf("NEGATIVE CYCLE\n");
    else
    {
        for(int i=0; i<nv; i++)
        {
            if(g.d[i]==INF) printf("INF\n");
            else printf("%d\n", g.d[i]);
        }
    }
    
    return 0;
}