Problem: For a given weighted graph G(V, E) and a source r, find the source shortest path to each vertex from the source (SSSP: Single Source Shortest Path).

Constraints

* 1 ≤ |V| ≤ 100000
* 0 ≤ di ≤ 10000
* 0 ≤ |E| ≤ 500000
* 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

Print the costs of SSSP in the following format.


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 1
0 2 4
1 2 2
2 3 1
1 3 5


Output

0
1
3
4


Input

4 6 1
0 1 1
0 2 4
2 0 1
1 2 2
3 1 1
3 2 5


Output

3
0
2
INF

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long INF = 1_000_000_000_000_000_000L;
        int V = Integer.parseInt(sc.next());
        int E = Integer.parseInt(sc.next());
        int r = Integer.parseInt(sc.next());
        ArrayList<ArrayList<int[]>> edge = new ArrayList<ArrayList<int[]>>();
        for (int i=0;i<V;i++) {
            ArrayList<int[]> add = new ArrayList<int[]>();
            edge.add(add);
        }
        for (int i=0;i<E;i++) {
            int s = Integer.parseInt(sc.next());
            int t = Integer.parseInt(sc.next());
            int d = Integer.parseInt(sc.next());
            int[] add = {t, d};
            edge.get(s).add(add);
        }

        long[] dist = new long[V];
        for (int i=0;i<V;i++) {
            dist[i] = INF;
        }

        PriorityQueue<long[]> q = new PriorityQueue<long[]>((o1, o2)->Long.compare(o1[1], o2[1]));
        long[] first = {r, 0};
        q.add(first);
        while (!q.isEmpty()) {
            long[] rem = q.poll();
            int index = (int)rem[0];
            long d = rem[1];

            dist[index] = Math.min(dist[index], d);
            for (int[] tmp : edge.get(index)) {
                long[] add = {tmp[0], tmp[1]+d};
                if (dist[tmp[0]]==INF) q.add(add);
            }
        }

        for (int i=0;i<V;i++) {
            if (dist[i]==INF) System.out.println("INF");
            else System.out.println(dist[i]);
        }
    }
}