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;
const int N = 200100;
int n, k;
int u[N], d[N], s[N];
vector<vector<int>> cnt;
vector<int> g[N];
long long current = 0;
long long ans = 0;
void dfs(int v) {
  if (u[v]) return;
  u[v] = s[v] = 1;
  for (auto i : g[v]) {
    if (!u[i]) {
      d[i] = d[v] + 1;
      dfs(i);
      cnt[v][1 % k]++;
      for (int j = 0; j < k; ++j) {
        cnt[v][(j + 1) % k] += cnt[i][j];
      }
      s[v] += s[i];
    }
  }
}
void calc(int v, long long current, vector<int> another) {
  if (u[v]) return;
  ans += current;
  u[v] = 1;
  vector<int> kek = another;
  for (int i = 0; i < k; ++i) {
    another[(i + 1) % k] = kek[i];
  }
  ++another[1 % k];
  for (auto i : g[v]) {
    if (!u[i]) {
      ++another[2 % k];
      for (int j = 0; j < k; ++j) {
        another[(j + 2) % k] += cnt[i][j];
      }
    }
  }
  for (auto i : g[v]) {
    if (!u[i]) {
      long long ncur = current;
      ncur -= cnt[i][0] + 1;
      --another[2 % k];
      for (int j = 0; j < k; ++j) {
        another[(j + 2) % k] -= cnt[i][j];
      }
      if (k == 1)
        ncur += another[0];
      else
        ncur += another[1];
      calc(i, ncur, another);
      ++another[2 % k];
      for (int j = 0; j < k; ++j) {
        another[(j + 2) % k] += cnt[i][j];
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cnt.resize(N, vector<int>(5));
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1);
  for (int i = 1; i <= n; ++i) {
    u[i] = 0;
    current += (d[i] / k + (d[i] % k != 0));
  }
  vector<int> a = {0, 0, 0, 0, 0};
  calc(1, current, a);
  cout << ans / 2LL << '\n';
}
