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;

using ll= long long;
const ll INF=1LL<<60;
const ll MOD=1000000007;

struct Edge {
    long long from;
    long long to;
    long long cost;
};
using Edges = vector<Edge>;


bool bellman_ford(Edges &Es, int V, int r, vector<ll> &dist){
    dist[r]=0;
    int cnt=0;
    while(cnt < V){
        bool end=true;
        for(auto x: Es){
            if(dist[x.from] != INF && dist[x.to] > dist[x.from]+x.cost){
                dist[x.to]=dist[x.from]+x.cost;
                end=false;
            }
        }
        if(end) break;
        cnt++;
    }
    return (cnt == V);
}

int main(){
    int V, E, r;
    Edges Es;
    vector<ll> dist;
    cin >> V >> E >> r;
    Es.resize(E);
    dist.assign(V, INF);
    for(int i=0; i<E; i++){
        int s, t, d;
        cin >> s >> t >> d;
        Es[i].from=s;
        Es[i].to=t;
        Es[i].cost=d;
    }
    if(bellman_ford(Es, V, r, dist)) cout << "NEGATIVE CYCLE" << endl;
    else{
        for(auto x: dist){
            if(x == INF) cout << "INF" << endl;
            else cout << x << endl;
        }
    }
    return 0;
}