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.
#include <bits/stdc++.h>
using namespace std;
int k;
vector<int> edge[300001];
long long ct[300001];
long long dp[300001];
long long ct1[300001][5];
long long an[5];
long long ans = 0;
void dfs(int v, int p) {
  dp[v] = 0;
  ct[v] = 1;
  for (auto u : edge[v]) {
    if (u != p) {
      dfs(u, v);
      dp[v] += dp[u] + ct[u];
      ct[v] += ct[u];
    }
  }
  for (auto u : edge[v]) {
    if (u != p) ans += (dp[u] + ct[u]) * (ct[v] - ct[u]);
  }
}
void dfs1(int v, int p) {
  ct1[v][0] = 1;
  for (auto u : edge[v]) {
    if (u != p) {
      dfs1(u, v);
      for (int c = 0; c < k; c++) {
        ct1[v][c] += ct1[u][(c + 1) % k];
      }
    }
  }
  for (int c = 0; c < k; c++) {
    an[c] += ct1[v][c] - (c == 0);
  }
  int ct[5] = {0, 0, 0, 0, 0};
  for (auto u : edge[v]) {
    if (u != p) {
      for (int c = 0; c < k; c++)
        for (int d = 0; d < k; d++)
          an[c] += ct1[u][d] * ct[(c - d + k + 2) % k];
      for (int d = 0; d < k; d++) ct[d] += ct1[u][d];
    }
  }
}
int main() {
  int n;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  dfs(0, 0);
  dfs1(0, 0);
  for (int c = 0; c < k; c++) {
    ans += an[c] * c;
  }
  cout << ans / k;
  return 0;
}
