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<utility>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<numeric>
#include<cassert>
#include<random>
using namespace std;
#define MK make_pair
#define F first
#define S second
#define ege pair<int,pair<int,int>> //w,s,t
const int MX=1e9+7;






int main(){
    int v,e,r;
    vector<ege> E;
    int dis[10000];
    fill(dis,&dis[10000],MX);
    cin>>v>>e>>r;
    dis[r]=0;
    for(int i=0;i<e;i++){
        int s,t,d;
        cin>>s>>t>>d;
        E.push_back({d,{s,t}});
    }
    for(int i=0;i<e;i++){
        for(int t=0;t<e;t++){
            ege s=E[t];
            if(dis[s.S.F]==MX){continue;}
            dis[s.S.S]=min(dis[s.S.S],dis[s.S.F]+s.F);
        }
    }
    bool j=false;
    for(int t=0;t<e;t++){
        ege s=E[t];
        if(dis[s.S.F]==MX){continue;}
        if(dis[s.S.S]>dis[s.S.F]+s.F){j=true; break;}
    }
    if(j){cout<<"NEGATIVE CYCLE"<<endl;}
    else{
        for(int i=0;i<v;i++){
            if(dis[i]==MX){cout<<"INF"<<endl; continue;}
            cout<<dis[i]<<endl;
        }
    }
    
    return 0;
}
