#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 5;
vector<int> adj[MAX], centroid_tree[MAX];
bool computed[MAX];
int size[MAX], centroid_father[MAX], dis[MAX];
int n, k, tree_size, centroid_root;
int dfs(int source, int father = -1, int cur_dis = 0) {
  dis[source] = cur_dis;
  size[source] = 1;
  tree_size++;
  for (auto &each : adj[source]) {
    if (each != father && !computed[each]) {
      size[source] += dfs(each, source, cur_dis + 1);
    }
  }
  return size[source];
}
int find_centroid(int source, int father = -1) {
  for (auto &each : adj[source]) {
    if (each != father && !computed[each] && size[each] > tree_size / 2) {
      return find_centroid(each, source);
    }
  }
  return source;
}
void build_centroid(int source, int last_centroid = -1) {
  tree_size = 0;
  dfs(source);
  int centroid = find_centroid(source);
  if (last_centroid != -1) {
    centroid_tree[last_centroid].push_back(centroid);
  } else {
    centroid_root = centroid;
  }
  centroid_father[centroid] = last_centroid;
  computed[centroid] = true;
  for (auto &each : adj[centroid]) {
    if (!computed[each]) {
      build_centroid(each, centroid);
    }
  }
}
long long get(int source, const vector<int> &app, const vector<long long> &len,
              vector<int> &new_app, vector<long long> &new_len) {
  long long v = (dis[source] + (k - 1)) / k;
  int r = dis[source] % k;
  long long total = v;
  for (int i = int(0); i < int(min(k, k + 1 - r)); i++) {
    total += v * app[i];
    if (r == 0 && i > 0) {
      total += app[i];
    }
    total += len[i];
  }
  for (int i = int(k + 1 - r); i < int(k); i++) {
    total += v * app[i];
    total += len[i];
    total += app[i];
  }
  for (auto &each : centroid_tree[source]) {
    total += get(each, app, len, new_app, new_len);
  }
  new_app[r]++;
  new_len[r] += dis[source] / k;
  return total;
}
long long roll(int source) {
  dfs(source);
  long long total = 0;
  vector<int> app(5, 0);
  vector<long long> len(5, 0);
  for (auto &each : centroid_tree[source]) {
    vector<int> new_app(5, 0);
    vector<long long> new_len(5, 0);
    total += get(each, app, len, new_app, new_len);
    for (int i = int(0); i < int(5); i++) {
      app[i] += new_app[i];
      len[i] += new_len[i];
    }
  }
  computed[source] = true;
  for (auto &each : centroid_tree[source]) {
    total += roll(each);
  }
  return total;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = int(0); i < int(n - 1); i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  build_centroid(1);
  memset(computed, false, sizeof computed);
  printf("%lld\n", roll(centroid_root));
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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 MAX = 2e5 + 5;
vector<int> adj[MAX], centroid_tree[MAX];
bool computed[MAX];
int size[MAX], centroid_father[MAX], dis[MAX];
int n, k, tree_size, centroid_root;
int dfs(int source, int father = -1, int cur_dis = 0) {
  dis[source] = cur_dis;
  size[source] = 1;
  tree_size++;
  for (auto &each : adj[source]) {
    if (each != father && !computed[each]) {
      size[source] += dfs(each, source, cur_dis + 1);
    }
  }
  return size[source];
}
int find_centroid(int source, int father = -1) {
  for (auto &each : adj[source]) {
    if (each != father && !computed[each] && size[each] > tree_size / 2) {
      return find_centroid(each, source);
    }
  }
  return source;
}
void build_centroid(int source, int last_centroid = -1) {
  tree_size = 0;
  dfs(source);
  int centroid = find_centroid(source);
  if (last_centroid != -1) {
    centroid_tree[last_centroid].push_back(centroid);
  } else {
    centroid_root = centroid;
  }
  centroid_father[centroid] = last_centroid;
  computed[centroid] = true;
  for (auto &each : adj[centroid]) {
    if (!computed[each]) {
      build_centroid(each, centroid);
    }
  }
}
long long get(int source, const vector<int> &app, const vector<long long> &len,
              vector<int> &new_app, vector<long long> &new_len) {
  long long v = (dis[source] + (k - 1)) / k;
  int r = dis[source] % k;
  long long total = v;
  for (int i = int(0); i < int(min(k, k + 1 - r)); i++) {
    total += v * app[i];
    if (r == 0 && i > 0) {
      total += app[i];
    }
    total += len[i];
  }
  for (int i = int(k + 1 - r); i < int(k); i++) {
    total += v * app[i];
    total += len[i];
    total += app[i];
  }
  for (auto &each : centroid_tree[source]) {
    total += get(each, app, len, new_app, new_len);
  }
  new_app[r]++;
  new_len[r] += dis[source] / k;
  return total;
}
long long roll(int source) {
  dfs(source);
  long long total = 0;
  vector<int> app(5, 0);
  vector<long long> len(5, 0);
  for (auto &each : centroid_tree[source]) {
    vector<int> new_app(5, 0);
    vector<long long> new_len(5, 0);
    total += get(each, app, len, new_app, new_len);
    for (int i = int(0); i < int(5); i++) {
      app[i] += new_app[i];
      len[i] += new_len[i];
    }
  }
  computed[source] = true;
  for (auto &each : centroid_tree[source]) {
    total += roll(each);
  }
  return total;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = int(0); i < int(n - 1); i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  build_centroid(1);
  memset(computed, false, sizeof computed);
  printf("%lld\n", roll(centroid_root));
  return 0;
}

```