A tree is an undirected connected graph without cycles. The distance between two vertices is the number of edges in a simple path between them.

Limak is a little polar bear. He lives in a tree that consists of n vertices, numbered 1 through n.

Limak recently learned how to jump. He can jump from a vertex to any vertex within distance at most k.

For a pair of vertices (s, t) we define f(s, t) as the minimum number of jumps Limak needs to get from s to t. Your task is to find the sum of f(s, t) over all pairs of vertices (s, t) such that s < t.


-----Input-----

The first line of the input contains two integers n and k (2 ≤ n ≤ 200 000, 1 ≤ k ≤ 5) — the number of vertices in the tree and the maximum allowed jump distance respectively.

The next n - 1 lines describe edges in the tree. The i-th of those lines contains two integers a_{i} and b_{i} (1 ≤ a_{i}, b_{i} ≤ n) — the indices on vertices connected with i-th edge.

It's guaranteed that the given edges form a tree.


-----Output-----

Print one integer, denoting the sum of f(s, t) over all pairs of vertices (s, t) such that s < t.


-----Examples-----
Input
6 2
1 2
1 3
2 4
2 5
4 6

Output
20

Input
13 3
1 2
3 2
4 2
5 2
3 6
10 6
6 7
6 13
5 8
5 9
9 11
11 12

Output
114

Input
3 5
2 1
3 1

Output
3



-----Note-----

In the first sample, the given tree has 6 vertices and it's displayed on the drawing below. Limak can jump to any vertex within distance at most 2. For example, from the vertex 5 he can jump to any of vertices: 1, 2 and 4 (well, he can also jump to the vertex 5 itself). [Image] 

There are $\frac{n \cdot(n - 1)}{2} = 15$ pairs of vertices (s, t) such that s < t. For 5 of those pairs Limak would need two jumps: (1, 6), (3, 4), (3, 5), (3, 6), (5, 6). For other 10 pairs one jump is enough. So, the answer is 5·2 + 10·1 = 20.

In the third sample, Limak can jump between every two vertices directly. There are 3 pairs of vertices (s < t), so the answer is 3·1 = 3.

Source: codeforces | Difficulty: HARD
Tags: Tree algorithms

Dynamic programming

Graph traversal


Solutions:
import sys
input = sys.stdin.buffer.readline
from collections import deque
(n, k) = map(int, input().split())
tot_dist = 0
vis = [0] * (n + 1)
dist = [[[0] * 2 for j in range(k)] for i in range(n + 1)]
adj = [[] for i in range(n + 1)]
for i in range(n - 1):
	(u, v) = map(int, input().split())
	adj[u].append(v)
	adj[v].append(u)
s = deque([1])
while s:
	c = s[-1]
	if not vis[c]:
		vis[c] = 1
		for ne in adj[c]:
			if not vis[ne]:
				s.append(ne)
	else:
		tot = [0] * k
		sum_dist = 0
		pairable = 0
		for ne in adj[c]:
			tot_dist += pairable * sum([dist[ne][i][0] for i in range(k)]) + sum_dist * sum([dist[ne][i][1] for i in range(k)])
			for i in range(k):
				for j in range(k):
					tot_dist += (i + j + 2 + k - 1) // k * dist[ne][i][1] * tot[j]
			for i in range(k):
				tot[i] += dist[ne][i][1]
				pairable += dist[ne][i][1]
				sum_dist += dist[ne][i][0]
		for ne in adj[c]:
			for i in range(k):
				for j in range(2):
					dist[c][i][j] += dist[ne][(i + k - 1) % k][j]
			dist[c][0][0] += dist[ne][k - 1][1]
		dist[c][0][1] += 1
		for i in range(k):
			tot_dist += dist[c][i][0] + (i + k - 1) // k * dist[c][i][1]
		s.pop()
print(tot_dist)


import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
from types import GeneratorType
BUFSIZE = 8192

class FastIO(IOBase):
	newlines = 0

	def __init__(self, file):
		import os
		self.os = os
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = 'x' in file.mode or 'r' not in file.mode
		self.write = self.buffer.write if self.writable else None

	def read(self):
		while True:
			b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
			if not b:
				break
			ptr = self.buffer.tell()
			(self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
		self.newlines = 0
		return self.buffer.read()

	def readline(self):
		while self.newlines == 0:
			b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
			self.newlines = b.count(b'\n') + (not b)
			ptr = self.buffer.tell()
			(self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
		self.newlines -= 1
		return self.buffer.readline()

	def flush(self):
		if self.writable:
			self.os.write(self._fd, self.buffer.getvalue())
			(self.buffer.truncate(0), self.buffer.seek(0))

class IOWrapper(IOBase):

	def __init__(self, file):
		self.buffer = FastIO(file)
		self.flush = self.buffer.flush
		self.writable = self.buffer.writable
		self.write = lambda s: self.buffer.write(s.encode('ascii'))
		self.read = lambda : self.buffer.read().decode('ascii')
		self.readline = lambda : self.buffer.readline().decode('ascii')
(sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout))
input = lambda : sys.stdin.readline().rstrip('\r\n')
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string

def getInts():
	return [int(s) for s in input().split()]

def getInt():
	return int(input())

def getStrs():
	return [s for s in input().split()]

def getStr():
	return input()

def listStr():
	return list(input())

def getMat(n):
	return [getInts() for _ in range(n)]
MOD = 10 ** 9 + 7

def bootstrap(f, stack=[]):

	def wrappedfunc(*args, **kwargs):
		if stack:
			return f(*args, **kwargs)
		else:
			to = f(*args, **kwargs)
			while True:
				if type(to) is GeneratorType:
					stack.append(to)
					to = next(to)
				else:
					stack.pop()
					if not stack:
						break
					to = stack[-1].send(to)
			return to
	return wrappedfunc

def solve():
	(N, K) = getInts()
	graph = dd(set)
	for i in range(N - 1):
		(A, B) = getInts()
		graph[A].add(B)
		graph[B].add(A)
	dp_count = [[0 for j in range(5)] for i in range(N + 1)]
	dp_total = [0 for j in range(N + 1)]
	nonlocal ans
	ans = 0

	@bootstrap
	def dfs(node, parent, depth):
		nonlocal ans
		dp_count[node][depth % K] = 1
		dp_total[node] = 1
		for neigh in graph[node]:
			if neigh != parent:
				yield dfs(neigh, node, depth + 1)
				for i in range(K):
					for j in range(K):
						diff = (i + j - 2 * depth) % K
						req = -diff % K
						ans += req * dp_count[node][i] * dp_count[neigh][j]
				for i in range(K):
					dp_count[node][i] += dp_count[neigh][i]
				dp_total[node] += dp_total[neigh]
		ans += dp_total[node] * (N - dp_total[node])
		yield
	dfs(1, -1, 0)
	return ans // K
print(solve())