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;
vector<long long int> v[1000000];
long long int dp[1000000][6];
long long int ans[1000000][6];
long long int sum = 0;
long long int n, k;
long long int ext = 0;
long long int vis[1000000];
long long int dfs(long long int a) {
  vis[a] = 1;
  long long int i, j, s;
  s = 0;
  for (i = 0; i <= v[a].size() - 1; i++) {
    long long int p = v[a][i];
    if (vis[p] == 1) continue;
    s += dfs(p);
  }
  s++;
  sum += s * (n - s);
  return s;
}
void mk(long long int a) {
  vis[a] = 1;
  long long int i, j, s;
  for (i = 0; i <= v[a].size() - 1; i++) {
    long long int p = v[a][i];
    if (vis[p] == 1) continue;
    mk(p);
    for (j = 0; j < k; j++) {
      ans[a][j] += dp[p][j];
    }
  }
}
void ck(long long int a) {
  vis[a] = 1;
  long long int i, j, s;
  for (i = 0; i <= v[a].size() - 1; i++) {
    long long int p = v[a][i];
    if (vis[p] == 1) {
      continue;
    }
    ck(p);
    for (j = 0; j < k; j++) {
      long long int t = dp[p][j];
      long long int u = j + 1;
      u = u % k;
      dp[a][u] += t;
    }
    long long int u = 1 % k;
    dp[a][u]++;
  }
  return;
}
void tk(long long int a) {
  vis[a] = 1;
  long long int i, j, s, x, y;
  for (i = 0; i <= v[a].size() - 1; i++) {
    long long int p = v[a][i];
    if (vis[p] == 1) continue;
    tk(p);
    for (j = 0; j < k; j++) {
      for (s = 0; s < k; s++) {
        x = dp[p][s];
        y = ans[a][j] - dp[p][j];
        long long int q = (s + j + 2) % k;
        if (q != 0) q = k - q;
        ext += q * x * y;
      }
    }
    for (j = 0; j < k; j++) ans[a][j] -= dp[p][j];
  }
}
int main() {
  long long int i, j, p, x, y;
  cin >> n >> k;
  for (i = 1; i <= n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  x = dfs(1);
  for (i = 0; i <= n; i++) vis[i] = 0;
  ck(1);
  for (i = 1; i <= n; i++) {
    for (j = 0; j < k; j++) {
      if (j == 0)
        ext += dp[i][j] * j;
      else
        ext += dp[i][j] * (k - j);
    }
  }
  for (i = 1; i <= n; i++) {
    dp[i][0]++;
  }
  for (i = 1; i <= n; i++) vis[i] = 0;
  mk(1);
  for (i = 1; i <= n; i++) vis[i] = 0;
  tk(1);
  cout << (sum + ext) / k;
}
