#include <bits/stdc++.h>
using namespace std;
vector<int> q[200010];
int n, k, a, b;
long long dis[200010];
long long pre[200010][6], dp[200010][6];
long long ans[200010], sum;
void dfs(int x, int fa) {
  pre[x][0] = 1;
  for (int i = 0; i < q[x].size(); i++) {
    int to = q[x][i];
    if (to == fa) continue;
    dis[to] = dis[x] + 1;
    dfs(to, x);
    for (int j = 0; j < k; j++) pre[x][(j + 1) % k] += pre[to][j];
  }
}
void dfs2(int x, int fa) {
  for (int i = 0; i < q[x].size(); i++) {
    int to = q[x][i];
    if (to == fa) continue;
    int tmp[7];
    for (int j = 0; j < k; j++) tmp[j] = dp[x][j] - dp[to][(j - 1 + k) % k];
    for (int j = 0; j < k; j++) {
      dp[to][j] += tmp[(j - 1 + k) % k];
    }
    ans[to] = ans[x] + dp[x][0] - pre[to][k - 1] - pre[to][0];
    dfs2(to, x);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    q[a].push_back(b);
    q[b].push_back(a);
  }
  dfs(1, -1);
  for (int i = 1; i <= n; i++) {
    ans[1] += (int)ceil(dis[i] * 1.0 / k);
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < k; j++) dp[i][j] = pre[i][j];
  dfs2(1, -1);
  for (int i = 1; i <= n; i++) sum += ans[i];
  printf("%lld\n", sum / 2);
  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;
vector<int> q[200010];
int n, k, a, b;
long long dis[200010];
long long pre[200010][6], dp[200010][6];
long long ans[200010], sum;
void dfs(int x, int fa) {
  pre[x][0] = 1;
  for (int i = 0; i < q[x].size(); i++) {
    int to = q[x][i];
    if (to == fa) continue;
    dis[to] = dis[x] + 1;
    dfs(to, x);
    for (int j = 0; j < k; j++) pre[x][(j + 1) % k] += pre[to][j];
  }
}
void dfs2(int x, int fa) {
  for (int i = 0; i < q[x].size(); i++) {
    int to = q[x][i];
    if (to == fa) continue;
    int tmp[7];
    for (int j = 0; j < k; j++) tmp[j] = dp[x][j] - dp[to][(j - 1 + k) % k];
    for (int j = 0; j < k; j++) {
      dp[to][j] += tmp[(j - 1 + k) % k];
    }
    ans[to] = ans[x] + dp[x][0] - pre[to][k - 1] - pre[to][0];
    dfs2(to, x);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    q[a].push_back(b);
    q[b].push_back(a);
  }
  dfs(1, -1);
  for (int i = 1; i <= n; i++) {
    ans[1] += (int)ceil(dis[i] * 1.0 / k);
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < k; j++) dp[i][j] = pre[i][j];
  dfs2(1, -1);
  for (int i = 1; i <= n; i++) sum += ans[i];
  printf("%lld\n", sum / 2);
  return 0;
}

```