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 ai and bi (1 ≤ ai, bi ≤ 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 <image> 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.
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:
        # update dists via pairing paths at the current node
        tot =  [0]*k
        sum_dist = 0
        pairable = 0
        for ne in adj[c]:
            # direct jumps of exactly k
            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)])
            # extra from remainder mod 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]
            # update pairable nodes
            for i in range(k):
                tot[i]+= dist[ne][i][1]
                pairable += dist[ne][i][1]
                sum_dist += dist[ne][i][0]
        # update paths
        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
        # update dists from path directly to current node
        for i in range(k):
            tot_dist += dist[c][i][0] + (i+k-1)//k * dist[c][i][1]

        s.pop()

print(tot_dist)