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<iostream>

using namespace std;

typedef long long ll;

#define rep(i,n) for(int i=0;i<(n);i++)

#define MAX_E 2000
#define MAX_V 1000
#define INF (1e18)
struct edge{ll from,to,cost;};

edge es[MAX_E];

ll d[MAX_V];
ll V,E;

bool shortest_path(ll s) {
    fill(d,d+V,INF);

    d[s] = 0;

    rep(i,V) {
        bool update = false;
        rep(j,E) {
            edge e = es[j];
            if(d[e.from] != INF && d[e.to] > d[e.from] + e.cost) {
                d[e.to] = d[e.from] + e.cost;
                update = true;
            }
        }
        if( !update) break;
        if(update && i == V - 1) {
            return false;;
        }
    }
    return true;
}

ll r;
int main() {
    cin >> V >> E >> r;
    rep(i,E) {
        ll s,t,dd;
        cin >> s >> t >> dd;
        es[i] = (edge){s,t,dd};
    }

    if(shortest_path(r)) {
        rep(i,V) {
            if(d[i] < INF)
                cout << d[i] << endl;
            else 
                cout << "INF" << endl;
        }
    }else {
        cout << "NEGATIVE CYCLE" << endl;
    }

    return 0;
}
