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;

struct edge {
    int f, t, c;
    edge(int f = 0, int t = 0, int c = 0): f(f), t(t), c(c) {}
};

const int INF = 1e9;
const int MAX_V = 1010;
const int MAX_E = 2010;
int V, E, r, d[MAX_V];
edge es[MAX_E];

bool bellmanFord(int s) {
    fill(d, d + V, INF);
    d[s] = 0;

    for (int i = 0; i < V; i++) {
        bool update = false;
        for (int j = 0; j < E; j++) {
            edge e = es[j];
            if (d[e.f] != INF && d[e.t] > d[e.f] + e.c) {
                if (i == V - 1) return false;
                d[e.t] = d[e.f] + e.c;
                update = true;
            }
        }
        if (!update) break;
    }
    return true;
}

int main(void) {
    cin >> V >> E >> r;
    for (int i = 0; i < E; i++)
        cin >> es[i].f >> es[i].t >> es[i].c;
    
    if (!bellmanFord(r)) cout << "NEGATIVE CYCLE\n";
    else {
        for (int i = 0; i < V; i++) {
            if (d[i] == INF) cout << "INF\n";
            else cout << d[i] << endl;
        }
    }

    return 0;
}
