#include <bits/stdc++.h>
using namespace std;
const int N = 200001;
int n, k, dpth[N], sz[N], dp[N][6];
long long ans;
vector<vector<int> > g;
void dfs(int u, int p) {
  sz[u] = 1;
  dp[u][dpth[u] % k] = 1;
  for (int i = 0; i < g[u].size(); i++)
    if (g[u][i] != p) {
      dpth[g[u][i]] = dpth[u] + 1;
      dfs(g[u][i], u);
      sz[u] += sz[g[u][i]];
      for (int a = 0; a < k; a++) {
        for (int b = 0; b < k; b++) {
          int rem = ((a + b) % k - (2 * dpth[u]) % k + k) % k;
          if (rem < k)
            ans += (long long)(((k - rem) % k + k) % k) * dp[u][a] *
                   dp[g[u][i]][b];
        }
      }
      for (int a = 0; a < k; a++) dp[u][a] += dp[g[u][i]][a];
    }
  ans += (long long)sz[u] * (n - sz[u]);
}
int main() {
  std::ios::sync_with_stdio(false);
  scanf("%d%d", &n, &k);
  g.resize(n);
  for (int i = 1, a, b; i < n; i++) {
    scanf("%d%d", &a, &b);
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dpth[0] = 0;
  ans = 0;
  dfs(0, -1);
  printf("%lld\n", ans / k);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 200001;
int n, k, dpth[N], sz[N], dp[N][6];
long long ans;
vector<vector<int> > g;
void dfs(int u, int p) {
  sz[u] = 1;
  dp[u][dpth[u] % k] = 1;
  for (int i = 0; i < g[u].size(); i++)
    if (g[u][i] != p) {
      dpth[g[u][i]] = dpth[u] + 1;
      dfs(g[u][i], u);
      sz[u] += sz[g[u][i]];
      for (int a = 0; a < k; a++) {
        for (int b = 0; b < k; b++) {
          int rem = ((a + b) % k - (2 * dpth[u]) % k + k) % k;
          if (rem < k)
            ans += (long long)(((k - rem) % k + k) % k) * dp[u][a] *
                   dp[g[u][i]][b];
        }
      }
      for (int a = 0; a < k; a++) dp[u][a] += dp[g[u][i]][a];
    }
  ans += (long long)sz[u] * (n - sz[u]);
}
int main() {
  std::ios::sync_with_stdio(false);
  scanf("%d%d", &n, &k);
  g.resize(n);
  for (int i = 1, a, b; i < n; i++) {
    scanf("%d%d", &a, &b);
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dpth[0] = 0;
  ans = 0;
  dfs(0, -1);
  printf("%lld\n", ans / k);
  return 0;
}

```