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:
// http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=GRL_3_A
// Prim's algorithm using a priority queue
#include <iostream>
#include <vector>
#include <queue>
#include <climits>

using namespace std;

int main(int arg, char **argv)
{
        int V, E;
        cin >> V >> E;
        vector<vector<pair<int, int>>> A(V);
        for (int i = 0; i < E; i++) {
                int s, t, w;
                cin >> s >> t >> w;
                A[s].push_back({w, t});
                A[t].push_back({w, s});
        }
        vector<bool> U(V);
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
        q.push({0, 0});
        int sum = 0;
        while (!q.empty()) {
                auto c = q.top();
                q.pop();
                if (U[c.second])
                        continue;
                U[c.second] = true;
                sum += c.first;
                for (auto &a: A[c.second])
                        if (!U[a.second])
                                q.push(a);
        }
        cout << sum << endl;
        return 0;
}