p03915 CODE FESTIVAL 2016 Final - Zigzag MST_9792
We have a graph with N vertices, numbered 0 through N-1. Edges are yet to be added.

We will process Q queries to add edges. In the i-th (1≦i≦Q) query, three integers A_i, B_i and C_i will be given, and we will add infinitely many edges to the graph as follows:

* The two vertices numbered A_i and B_i will be connected by an edge with a weight of C_i.
* The two vertices numbered B_i and A_i+1 will be connected by an edge with a weight of C_i+1.
* The two vertices numbered A_i+1 and B_i+1 will be connected by an edge with a weight of C_i+2.
* The two vertices numbered B_i+1 and A_i+2 will be connected by an edge with a weight of C_i+3.
* The two vertices numbered A_i+2 and B_i+2 will be connected by an edge with a weight of C_i+4.
* The two vertices numbered B_i+2 and A_i+3 will be connected by an edge with a weight of C_i+5.
* The two vertices numbered A_i+3 and B_i+3 will be connected by an edge with a weight of C_i+6.
* ...



Here, consider the indices of the vertices modulo N. For example, the vertice numbered N is the one numbered 0, and the vertice numbered 2N-1 is the one numbered N-1.

The figure below shows the first seven edges added when N=16, A_i=7, B_i=14, C_i=1:

<image>

After processing all the queries, find the total weight of the edges contained in a minimum spanning tree of the graph.

Constraints

* 2≦N≦200,000
* 1≦Q≦200,000
* 0≦A_i,B_i≦N-1
* 1≦C_i≦10^9

Input

The input is given from Standard Input in the following format:


N Q
A_1 B_1 C_1
A_2 B_2 C_2
:
A_Q B_Q C_Q


Output

Print the total weight of the edges contained in a minimum spanning tree of the graph.

Examples

Input

7 1
5 2 1


Output

21


Input

2 1
0 0 1000000000


Output

1000000001


Input

5 3
0 1 10
0 2 10
0 4 10


Output

42
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

from operator import itemgetter

N,Q = map(int,readline().split())
m = map(int,read().split())
ABC = list(zip(m,m,m))

INF = 10 ** 18
cyclic_cost = [INF] * N
for a,b,c in ABC:
    if cyclic_cost[a] > c + 1:
        cyclic_cost[a] = c + 1
    if cyclic_cost[b] > c + 2:
        cyclic_cost[b] = c + 2

cyclic_cost += cyclic_cost # 2周分
x = INF
for i in range(N+N):
    x += 2
    if x < cyclic_cost[i]:
        cyclic_cost[i] = x
    x = cyclic_cost[i]
cyclic_cost = [x if x < y else y for x,y in zip(cyclic_cost, cyclic_cost[N:])]

ABC += [(i,i+1,c) for i,c in enumerate(cyclic_cost)]
ABC[-1] = (N-1,0,cyclic_cost[-1])

class UnionFind:
    def __init__(self,N):
        self.root = list(range(N))
        self.size = [1] * (N)
        
    def find_root(self,x):
        root = self.root
        while root[x] != x:
            root[x] = root[root[x]]
            x = root[x]
        return x
    
    def merge(self,x,y):
        x = self.find_root(x)
        y = self.find_root(y)
        if x == y:
            return False
        sx,sy = self.size[x],self.size[y]
        if sx < sy:
            self.root[x] = y
            self.size[y] += sx
        else:
            self.root[y] = x
            self.size[x] += sy
        return True

ABC.sort(key = itemgetter(2))

uf = UnionFind(N)
merge = uf.merge
answer = 0
for a,b,c in ABC:
    if merge(a,b):
        answer += c

print(answer)
7 1
5 2 1 2 1
0 0 1000000000 5 3
0 1 10
0 2 10
0 4 10 7 1
6 2 1 2 1
0 0 1000000010 5 3
0 0 10
0 2 10
0 4 10 9 1
6 2 1 5 3
0 0 10
0 3 10
0 4 10 9 1
7 2 1 9 1
5 2 1 7 1
5 4 1 2 1
0 0 0000000000 5 3
0 1 10
1 2 10
0 4 10 2 1
1 0 1000000010 9 1
6 2 0 2 1
0 0 0000000010 5 3
0 0 10
1 2 10
0 4 16 5 3
0 0 10
1 0 10
0 4 16 2 1
0 0 1000000100 5 3
0 1 10
0 2 7
0 4 10 2 1
0 0 1000001010 5 3
0 0 10
0 3 10
0 4 3 9 1
6 1 0 2 1
0 0 0000010010 9 1
1 2 0 12 1
3 2 1 2 1
0 0 1000010100 7 1
2 4 1 3 1
0 0 1000001010 2 1
0 0 0001010010 9 1
2 2 0 5 3
0 0 10
1 0 10
0 4 0 12 1
3 4 1 2 1
0 0 1000010110 3 1
1 0 1000001010 13 1
6 1 -1 5 3
0 0 10
0 2 10
1 4 2 12 1
3 4 2 2 1
0 0 1000010010 12 1
6 1 -1 21 1
3 4 2 12 1
2 1 -1 21 1
3 4 0 5 1
2 1 -1 21 1
3 4 1 5 1
1 1 -1 21 1
1 4 1 5 1
0 1 -1 21 1
0 4 1 9 1
0 1 -1 9 1
0 2 -1 6 1
5 2 1 2 1
0 0 1000100000 7 1
6 2 0 2 1
0 0 1000100010 9 1
6 4 1 5 3
0 1 10
1 2 10
0 4 9 4 1
1 0 1000000010 2 1
0 1 0000000010 12 1
5 2 0 7 1
2 0 1 2 1
0 0 1000001000 16 1
6 1 0 2 1
0 0 0000010011 4 1
0 0 1000001010 9 1
6 2 -1 12 1
6 4 1 4 1
1 0 1000001010 5 3
1 0 2
1 2 20
0 4 10 13 1
6 1 0 7 1
3 2 0 12 1
1 4 2 21 1
6 4 2 21 1
0 4 0 7 1
1 1 -1 21 1
1 0 1 2 1
1 1 -1 32 1
0 4 1 2 1
0 0 1000110010 9 1
6 4 2 3 1
1 0 1000000010 2 1
1 0 1000001000 5 3
0 0 10
0 0 10
1 4 3 8 3
0 0 10
1 2 10
0 4 10 5 3
1 0 14
1 2 10
0 4 21 12 1
0 2 0 9 1
6 6 1 5 3
1 0 1
1 2 20
0 4 10 13 1
6 1 1 12 1
1 7 2 21 1
6 4 1 21 1
1 4 0 32 1
0 5 1 3 1
2 0 1000000010 14 3
0 0 10
1 2 10
0 4 10 9 1
2 0 0 13 1
6 0 1 12 1
2 7 2 28 1
6 4 1 34 1
1 4 0 29 1
0 0 1 4 1
2 0 1000000010 2 1
1 1 0000011011
21 1000000001 42 22
 1000000011
 43
 36
 44
 37
 39
 31
 1
 42
 1000000010
 28
 11
 47
 49
 1000000101
 34
 1000001011
 21
 29
 10011
 56
 111
 1000010101
 27
 2000002024
 1010011
 64
 9
 121
 1000010111
 2000002021
 57
 14
 132
 1000010011
 45
 420
 89
 380
 5
 400
 12
 330
 8
 301
 48
 35
 15
 1000100001
 16
 1000100011
 46
 41
 3000000034
 10
 70
 24
 1000001001
 120
 10012
 3000003039
 20
 94
 3000003034
 17
 69
 25
 98
 366
 281
 30
 381
 0
 796
 1000110011
 54
 2000000021
 1000001000
 18
 85
 50
 91
 72
 13
 81
 77
 346
 310
 749
 2000000022
 223
 38
 78
 80
 654
 934
 812
 3000000033
 11012
