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;
using ll = long long;
const ll INF = 1e18;

int main()
{
    ll n, m, r;
    cin >> n >> m >> r;
    vector<vector<pair<ll,ll>>> hen(n);
    vector<ll> dist(n, INF);
    for(ll i = 0; i < m; i++) {
        ll a, b, d;
        cin >> a >> b >> d;
        hen[a].emplace_back(b, d);
    }

    {
        bool ok = true;
        ll cycle = 0;
        dist[r] = 0;
        while(ok) {
            ok = false;
            for(ll i = 0; i < n; i++) {
                if(dist[i] == INF)continue;
                for(auto j : hen[i]) {
                    if(dist[j.first] > dist[i] + j.second)ok = true;
                    dist[j.first] = min(dist[j.first], dist[i] + j.second);
                }
            }
            if(cycle >= 2 * n && ok) {
                cout << "NEGATIVE CYCLE" << endl;
                return 0;
            }
            cycle++;
        }
    }

    for(ll i = 0; i < n; i++) {
        if(dist[i] == INF) {
            cout << "INF" << endl;
        } else {
            cout << dist[i] << endl;
        }
    }


}
