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 long long mx = 300000;
long long n, k, dp[mx][5], dp2[mx][5], p[mx], ans, sz[mx];
vector<int> g[mx], ch[mx];
void dfs(int v, int pv) {
  p[v] = pv;
  if (pv != -1) ch[pv].push_back(v);
  for (int i = 0; i < (int)g[v].size(); ++i) {
    if (g[v][i] != pv) dfs(g[v][i], v);
  }
}
void count_dp1(int v) {
  sz[v] = 0;
  for (int i = 0; i < k; ++i) {
    dp[v][i] = 0;
  }
  for (int i = 0; i < (int)ch[v].size(); ++i) {
    int v1 = ch[v][i];
    count_dp1(v1);
    for (int j = 0; j < k; ++j) {
      dp[v][j] += dp[v1][j];
    }
    sz[v] += sz[v1];
  }
  long long temp[5];
  for (int i = 0; i < k; ++i) {
    temp[(i + 1) % k] = dp[v][i];
  }
  temp[0]++;
  for (int i = 0; i < k; ++i) {
    dp[v][i] = temp[i];
  }
  sz[v]++;
}
void count_dp2(int v) {
  for (int j = 0; j < (int)ch[v].size(); ++j) {
    int v1 = ch[v][j];
    long long temp[5];
    for (int i = 0; i < 5; ++i) {
      temp[i] = 0;
    }
    for (int i = 0; i < k; ++i) {
      temp[(i + 1) % k] = dp[v][i];
    }
    for (int i = 0; i < k; ++i) {
      temp[(i + 2) % k] -= dp[v1][i];
    }
    for (int i = 0; i < k; ++i) {
      temp[(i + 1) % k] += dp2[v][i];
    }
    for (int i = 0; i < k; ++i) {
      dp2[v1][i] = temp[i];
    }
  }
  for (int j = 0; j < (int)ch[v].size(); ++j) {
    int v1 = ch[v][j];
    count_dp2(v1);
  }
}
void dfs3(int v) {
  if (v != 0) {
    ans += (dp[v][0] * (n - sz[v]) + dp2[v][1 % k] * (sz[v]));
  }
  for (int i = 0; i < (int)ch[v].size(); ++i) {
    dfs3(ch[v][i]);
  }
}
int main() {
  ans = 0;
  int u, v;
  cin >> n >> k;
  for (long long i = 0; i < (n - 1); ++i) {
    cin >> u >> v;
    g[u - 1].push_back(v - 1);
    g[v - 1].push_back(u - 1);
  }
  dfs(0, -1);
  count_dp1(0);
  for (int i = 0; i < 5; ++i) {
    dp2[0][i] = 0;
  }
  count_dp2(0);
  dfs3(0);
  cout << ans / 2 << endl;
  return 0;
}
