#include <bits/stdc++.h>
using namespace std;
const int sz = 2e5;
long long tot;
long long n, k;
bool vis[sz];
vector<vector<int> > g;
long long depsum[sz];
long long up[5][sz], down[5][sz], subsz[sz];
int par[sz];
long long tot2[5];
void dfs1(int u) {
  vis[u] = true;
  subsz[u] = 1;
  for (int i = 0; i < k; ++i) up[i][u] = 0;
  up[0][u] = 1;
  for (int v : g[u])
    if (!vis[v]) {
      par[v] = u;
      dfs1(v);
      tot += subsz[v] * (n - subsz[v]);
      subsz[u] += subsz[v];
      for (int i = 0; i < k; ++i) up[(i + 1) % k][u] += up[i][v];
    }
}
void dfs2(int u) {
  if (par[u] == u)
    for (int i = 0; i < k; ++i) down[i][u] = 0;
  for (int i = 0; i < k; ++i) {
    tot2[i] += down[i][u] + up[i][u];
  }
  tot2[0]--;
  for (int v : g[u])
    if (v != par[u])
      for (int i = 0; i < k; ++i)
        down[(i + 2) % k][v] +=
            down[(i + 1) % k][u] + up[(i + 1) % k][u] - up[i][v];
  for (int v : g[u])
    if (v != par[u]) dfs2(v);
}
int main() {
  scanf("%d%d", &n, &k);
  g.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < n; ++i) vis[i] = false;
  tot = 0;
  dfs1(0);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < k; ++j) down[j][i] = 0;
  for (int i = 0; i < k; ++i) tot2[i] = 0;
  dfs2(0);
  long long ans = tot + (k - 1) * n * (n - 1) / 2;
  for (int i = 0; i < k; ++i) {
    ans -= ((k + i - 1) % k) * tot2[i] / 2;
  }
  ans /= k;
  cout << ans << endl;
}

### Prompt

Create a solution in Cpp for the following problem:

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 sz = 2e5;
long long tot;
long long n, k;
bool vis[sz];
vector<vector<int> > g;
long long depsum[sz];
long long up[5][sz], down[5][sz], subsz[sz];
int par[sz];
long long tot2[5];
void dfs1(int u) {
  vis[u] = true;
  subsz[u] = 1;
  for (int i = 0; i < k; ++i) up[i][u] = 0;
  up[0][u] = 1;
  for (int v : g[u])
    if (!vis[v]) {
      par[v] = u;
      dfs1(v);
      tot += subsz[v] * (n - subsz[v]);
      subsz[u] += subsz[v];
      for (int i = 0; i < k; ++i) up[(i + 1) % k][u] += up[i][v];
    }
}
void dfs2(int u) {
  if (par[u] == u)
    for (int i = 0; i < k; ++i) down[i][u] = 0;
  for (int i = 0; i < k; ++i) {
    tot2[i] += down[i][u] + up[i][u];
  }
  tot2[0]--;
  for (int v : g[u])
    if (v != par[u])
      for (int i = 0; i < k; ++i)
        down[(i + 2) % k][v] +=
            down[(i + 1) % k][u] + up[(i + 1) % k][u] - up[i][v];
  for (int v : g[u])
    if (v != par[u]) dfs2(v);
}
int main() {
  scanf("%d%d", &n, &k);
  g.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < n; ++i) vis[i] = false;
  tot = 0;
  dfs1(0);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < k; ++j) down[j][i] = 0;
  for (int i = 0; i < k; ++i) tot2[i] = 0;
  dfs2(0);
  long long ans = tot + (k - 1) * n * (n - 1) / 2;
  for (int i = 0; i < k; ++i) {
    ans -= ((k + i - 1) % k) * tot2[i] / 2;
  }
  ans /= k;
  cout << ans << endl;
}

```