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<bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i=0; i<n; i++)
#define INF 1000000000

typedef pair<int , int > P;

struct edge{
  int to, cost;
  edge(int to, int cost):
    to(to), cost(cost){}
};
struct vertex{
  vector<edge> E;
};

vertex V[100000];

int nv, ne;
int r;
int d[100000];

void dijkstra(int s){
  //?????????????????? first:???????????¢ second:??????
  priority_queue<P, vector<P>, greater<P> > que;
  fill(d, d+nv, INF);
  d[s]=0;
  que.push(P(0, s));

  while(!que.empty()){
    P p=que.top(); que.pop();
    int v=p.second;
    if(d[v]<p.first)continue;
    rep(i, V[v].E.size()){
      edge e=V[v].E[i];
      if(d[e.to]>d[v]+e.cost){
	d[e.to]=d[v]+e.cost;
	que.push(P(d[e.to], e.to));
      }
    }
  }
}

int main(){
  cin>>nv>>ne>>r;
  rep(i, ne){
    int s, d, c;
    cin>>s>>d>>c;
    V[s].E.push_back(edge(d, c));
  }
  dijkstra(r);
  rep(i,nv)
    if(d[i]==INF)cout<<"INF"<<endl;
    else cout<<d[i]<<endl;

  return 0;
}