#include <bits/stdc++.h>
using namespace std;
struct NODE {
  int to, next;
} node[400000 + 10];
int n, k, tol;
long long ans;
long long dp[200000 + 10][10], sum[200000 + 10];
int head[200000 + 10];
void addedge(int from, int to) {
  node[tol].to = to;
  node[tol].next = head[from];
  head[from] = tol++;
}
void tree_dp(int now, int fa, int depth) {
  dp[now][depth % k] = sum[now] = 1;
  for (int i = head[now]; i != -1; i = node[i].next) {
    int to = node[i].to;
    if (to != fa) {
      tree_dp(to, now, depth + 1);
      for (int a = 0; a < k; a++) {
        for (int b = 0; b < k; b++) {
          int t = ((a + b) % k - (2 * depth) % k + k) % k;
          t = (k - t + k) % k;
          ans += t * dp[now][a] * dp[to][b];
        }
      }
      for (int a = 0; a < k; a++) dp[now][a] += dp[to][a];
      sum[now] += sum[to];
      ans += sum[to] * (n - sum[to]);
    }
  }
}
int main() {
  while (scanf("%d %d", &n, &k) != EOF) {
    memset(head, -1, sizeof(head));
    tol = 0;
    ans = 0ll;
    for (int i = 1; i <= n - 1; i++) {
      int from, to;
      scanf("%d %d", &from, &to);
      addedge(from, to);
      addedge(to, from);
    }
    memset(dp, 0, sizeof(dp));
    memset(sum, 0, sizeof(sum));
    tree_dp(1, 0, 0);
    printf("%lld\n", ans / k);
  }
  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;
struct NODE {
  int to, next;
} node[400000 + 10];
int n, k, tol;
long long ans;
long long dp[200000 + 10][10], sum[200000 + 10];
int head[200000 + 10];
void addedge(int from, int to) {
  node[tol].to = to;
  node[tol].next = head[from];
  head[from] = tol++;
}
void tree_dp(int now, int fa, int depth) {
  dp[now][depth % k] = sum[now] = 1;
  for (int i = head[now]; i != -1; i = node[i].next) {
    int to = node[i].to;
    if (to != fa) {
      tree_dp(to, now, depth + 1);
      for (int a = 0; a < k; a++) {
        for (int b = 0; b < k; b++) {
          int t = ((a + b) % k - (2 * depth) % k + k) % k;
          t = (k - t + k) % k;
          ans += t * dp[now][a] * dp[to][b];
        }
      }
      for (int a = 0; a < k; a++) dp[now][a] += dp[to][a];
      sum[now] += sum[to];
      ans += sum[to] * (n - sum[to]);
    }
  }
}
int main() {
  while (scanf("%d %d", &n, &k) != EOF) {
    memset(head, -1, sizeof(head));
    tol = 0;
    ans = 0ll;
    for (int i = 1; i <= n - 1; i++) {
      int from, to;
      scanf("%d %d", &from, &to);
      addedge(from, to);
      addedge(to, from);
    }
    memset(dp, 0, sizeof(dp));
    memset(sum, 0, sizeof(sum));
    tree_dp(1, 0, 0);
    printf("%lld\n", ans / k);
  }
  return 0;
}

```