#include <bits/stdc++.h>
using namespace std;
struct Edge {
  int v, next;
} edge[200050 * 2];
int e, ft[200050];
int n, k;
int size[200050];
long long dp[200050][5 + 1];
bool flag[200050];
long long total_dist;
void init() {
  memset(flag, false, sizeof(flag));
  memset(dp, 0, sizeof(dp));
  memset(ft, -1, sizeof(ft));
  memset(size, 0, sizeof(size));
  e = 0;
}
void add_edge(int u, int v) {
  edge[e].v = v;
  edge[e].next = ft[u];
  ft[u] = e++;
}
void dfs(int u) {
  flag[u] = true;
  size[u] = 1;
  long long ans = 0;
  long long tmp[5 + 1];
  for (int i = 0; i <= k; i++) dp[u][i] = 0;
  dp[u][0] = 1;
  for (int i = ft[u]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    if (flag[v]) continue;
    dfs(v);
    tmp[k] = dp[v][k];
    for (int j = 0; j < k; j++) {
      tmp[(j + 1) % k] = dp[v][j];
      if (j + 1 == k) tmp[k] += dp[v][j];
    }
    for (int x = 0; x < k; x++) {
      for (int y = 0; y < k; y++) {
        long long d = (x + y + k - 1) / k;
        total_dist += d * dp[u][x] * tmp[y];
      }
    }
    total_dist += size[u] * tmp[k];
    total_dist += size[v] * dp[u][k];
    for (int j = 0; j <= k; j++) {
      dp[u][j] += tmp[j];
    }
    size[u] += size[v];
  }
}
int main() {
  int u, v;
  while (cin >> n >> k) {
    init();
    for (int i = 1; i < n; i++) {
      scanf("%d%d", &u, &v);
      add_edge(u, v);
      add_edge(v, u);
    }
    total_dist = 0;
    dfs(1);
    cout << total_dist << endl;
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
struct Edge {
  int v, next;
} edge[200050 * 2];
int e, ft[200050];
int n, k;
int size[200050];
long long dp[200050][5 + 1];
bool flag[200050];
long long total_dist;
void init() {
  memset(flag, false, sizeof(flag));
  memset(dp, 0, sizeof(dp));
  memset(ft, -1, sizeof(ft));
  memset(size, 0, sizeof(size));
  e = 0;
}
void add_edge(int u, int v) {
  edge[e].v = v;
  edge[e].next = ft[u];
  ft[u] = e++;
}
void dfs(int u) {
  flag[u] = true;
  size[u] = 1;
  long long ans = 0;
  long long tmp[5 + 1];
  for (int i = 0; i <= k; i++) dp[u][i] = 0;
  dp[u][0] = 1;
  for (int i = ft[u]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    if (flag[v]) continue;
    dfs(v);
    tmp[k] = dp[v][k];
    for (int j = 0; j < k; j++) {
      tmp[(j + 1) % k] = dp[v][j];
      if (j + 1 == k) tmp[k] += dp[v][j];
    }
    for (int x = 0; x < k; x++) {
      for (int y = 0; y < k; y++) {
        long long d = (x + y + k - 1) / k;
        total_dist += d * dp[u][x] * tmp[y];
      }
    }
    total_dist += size[u] * tmp[k];
    total_dist += size[v] * dp[u][k];
    for (int j = 0; j <= k; j++) {
      dp[u][j] += tmp[j];
    }
    size[u] += size[v];
  }
}
int main() {
  int u, v;
  while (cin >> n >> k) {
    init();
    for (int i = 1; i < n; i++) {
      scanf("%d%d", &u, &v);
      add_edge(u, v);
      add_edge(v, u);
    }
    total_dist = 0;
    dfs(1);
    cout << total_dist << endl;
  }
  return 0;
}

```