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
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <string.h>
#include <math.h>
#include <time.h>
#include <stdlib.h>
#include <queue>

#define INF 2147483647

using namespace std;

struct edge { int to,cost;};	
typedef pair<int,int> P;

int main(){
	int V,E,r,
	    s,t,d,
	    score[100000];
	P p;
	edge e;
	vector<edge> G[100000];
	priority_queue<P,vector<P>,greater<P> > q;
	
	scanf("%d %d %d",&V,&E,&r);
	for(int i=0;i<E;i++){
		scanf("%d %d %d",&s,&t,&d);
		G[s].push_back( {t,d} );
	}
	
	fill(score , score + V , INF);
	score[r] = 0;
	q.push( P(0,r) );
	
	while(!q.empty()){
		p = q.top();
		q.pop();
		r = p.second;
		if(score[r] < p.first) continue;
		for(int j=0;j<G[r].size();j++){
			e = G[r][j];
			if(score[e.to] > score[r]+e.cost){
				score[e.to] = score[r]+e.cost;
				q.push( P(score[e.to],e.to) );
			}
		}
	}
	for(int i=0;i<V;i++){
		if(score[i]==2147483647) printf("INF\n");
		else printf("%d\n",score[i]);
	}
		
	return 0;
}
