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):
        self.par = [i for i in range(n+1)]
        self.rank = [0] * (n+1)

    # 検索
    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]

    # 併合
    def union(self, x, y):
        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

    # 同じ集合に属するか判定
    def same_check(self, x, y):
        return self.find(x) == self.find(y)
V, E = map(int, input().split())
X = [list(map(int, input().split())) for _ in range(E)]
X = sorted(X, key = lambda x:x[2])
UF = UnionFind(V)
ans = 0
for i in range(E):
    a, b = X[i][0]+1, X[i][1]+1
    if not UF.same_check(a, b):
        UF.union(a, b)
        ans += X[i][2]
print(ans)