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 PYTHON3):

Solution:
class UnionFind:
    def __init__(self, n):
        #親要素のノード番号を格納。par[x] == xのときそのノードはroot
        self.par = [i for i in range(n)]
        #木の高さ（初期状態では0）
        self.rank = [0] * n

    def find(self, x):
        #木の根を求める
        if self.par[x] == x:
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]

    def same(self, x, y):
        # xとyが同じ集合に属するかを判定
        return self.find(x) == self.find(y)

    def union(self, x, y):
        #併合する
        #まず根を探す
        x = self.find(x)
        y = self.find(y)
        if self.rank[x] < self.rank[y]:
            self.par[x] = y
        else:
            self.par[y] = x
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1
V, E = map(int, input().split())
nodes = sorted([list(map(int, input().split())) for _ in range(E)], key = lambda x: x[2])

a = UnionFind(V)
ans = 0
for node in nodes:
    si, ti = node[0] - 1, node[1] - 1
    if not a.same(si, ti):
        a.union(si, ti)
        ans += node[2]

print(ans)