Problem: Find the sum of weights of edges of the Minimum Spanning Tree for a given weighted undirected graph G = (V, E).

Constraints

* 1 ≤ |V| ≤ 10,000
* 0 ≤ |E| ≤ 100,000
* 0 ≤ wi ≤ 10,000
* The graph is connected
* There are no parallel edges
* There are no self-loops

Input


|V| |E|
s0 t0 w0
s1 t1 w1
:
s|E|-1 t|E|-1 w|E|-1


, where |V| is the number of vertices and |E| is the number of edges in the graph. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively.

si and ti represent source and target verticess of i-th edge (undirected) and wi represents the weight of the i-th edge.

Output

Print the sum of the weights of the Minimum Spanning Tree.

Examples

Input

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


Output

3


Input

6 9
0 1 1
0 2 3
1 2 1
1 3 7
2 4 1
1 4 3
3 4 1
3 5 1
4 5 6


Output

5

Solution (in CPP):

Solution:
#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
typedef pair<int,int> PA;
int main(){
  int v,e,s,t,w;
  vector<PA> vec[100001];
  scanf("%d %d",&v,&e);
  for(int i=0;i<e;i++){
    scanf("%d %d %d",&s,&t,&w);
    vec[s].push_back(PA(w,t));
    vec[t].push_back(PA(w,s));
  }
  priority_queue<PA,vector<PA>,greater<PA> > pq;
  pq.push(PA(0,0));
  bool g[100001]={false};
  int cnt = 0;
  while(!pq.empty()){
    PA pt = pq.top();
    pq.pop();
    if(g[pt.second]) continue;
    g[pt.second] = true;
    cnt += pt.first;
    for(int i=0;i<vec[pt.second].size();i++){
      pq.push(vec[pt.second][i]);
    }
  }
  printf("%d\n",cnt);
  return(0);
}