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;
typedef struct{
    int from, to, cost;
}Edge;
const int INF = 0x3f3f3f3f;

int main(){
    int n, k, s;
    scanf("%d %d %d", &n, &k, &s);
    Edge e[k];
    int d[n];
    for(int i = 0 ; i < k; i++)
        scanf("%d %d %d", &e[i].from, &e[i].to, &e[i].cost);
    for(int i = 0; i < n; i++) d[i] = (i == s) ? 0 : INF;

    int cnt;
    for(cnt = 0 ; cnt < n; cnt++) {
        int update = 0;
        for (int i = 0; i < k; i++) {
            if (d[e[i].from] != INF && d[e[i].from] + e[i].cost < d[e[i].to]) {
                d[e[i].to] = d[e[i].from] + e[i].cost;
                update = 1;
            }
        }
        if (!update) break;
    }
    int flag = 0;
    if(cnt == n) flag = 1;
    if(flag){
        printf("NEGATIVE CYCLE\n");
    }else{
        for(int i = 0; i < n; i++){
            d[i] == INF ? printf("INF\n") : printf("%d\n", d[i]);
        }
    }
    return 0;
}
