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 <iostream>
#include <vector>
#include <queue>

using namespace std;

typedef pair<int , int> P;

int main(){
    int V, E;
    cin >> V >> E;

    vector<P> G[V];
    for(int i=0; i<E; i++){
        int s, t, w;
        cin >> s >> t >> w;
        G[s].push_back(make_pair(t, w));
        G[t].push_back(make_pair(s, w));
    }

    int ans = 0;
    priority_queue<P, vector<P>, greater<P>> q;
    bool used[V];
    fill(used, used+V, false);
    q.push(make_pair(0, 0));
    int cnt = 0;
    while(true){
        if(cnt==V) break;
        int now = q.top().second;
        int d = q.top().first;
        q.pop();
        if(used[now]) continue;

        used[now] = true;
        ans += d;
        cnt++;

        for(int i=0; i<G[now].size(); i++){
            int next = G[now][i].first;
            int c = G[now][i].second;
            q.push(make_pair(c, next));
        }
    }

    cout << ans << endl;

    return 0;
}