#include <bits/stdc++.h>
using namespace std;
int n, k;
vector<int> tree[200002];
long long int total = 0;
long long int dp[200002][6];
long long int modSum[6];
long long int dfs1(int nd, int p) {
  long long int sum = 1;
  for (int i = 0; i < (int)tree[nd].size(); i++) {
    int v = tree[nd][i];
    if (v == p) continue;
    long long int part = dfs1(v, nd);
    sum += part;
    total += part * (n - part);
  }
  return sum;
}
void dfs2(int nd, int p) {
  for (int i = 0; i < (int)tree[nd].size(); i++) {
    int v = tree[nd][i];
    if (v == p) continue;
    dfs2(v, nd);
    for (int x = 0; x < k; x++) {
      for (int y = 1; y <= k; y++) {
        int ym = (y % k);
        ym = (ym ? ym : k);
        int yy = ym - 1;
        int nm = (x + ym) % k;
        long long int a = dp[nd][x];
        long long int b = dp[v][yy];
        long long int c = (ym == (1 % k));
        modSum[nm] += (a * (b + c));
      }
    }
    dp[nd][1 % k]++;
    modSum[1 % k]++;
    for (int j = 0; j < k - 1; j++) {
      dp[nd][j + 1] += dp[v][j];
      modSum[j + 1] += dp[v][j];
    }
    dp[nd][0] += dp[v][k - 1];
    modSum[0] += dp[v][k - 1];
  }
}
int main() {
  scanf("%d %d", &n, &k);
  int a, b;
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &a, &b);
    tree[a].push_back(b);
    tree[b].push_back(a);
  }
  memset(modSum, 0, sizeof(modSum));
  memset(dp, 0, sizeof(dp));
  dfs1(1, 0);
  dfs2(1, 0);
  for (int i = 1; i < k; i++) {
    total += (modSum[i] * (k - i));
  }
  printf("%I64d", total / 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;
int n, k;
vector<int> tree[200002];
long long int total = 0;
long long int dp[200002][6];
long long int modSum[6];
long long int dfs1(int nd, int p) {
  long long int sum = 1;
  for (int i = 0; i < (int)tree[nd].size(); i++) {
    int v = tree[nd][i];
    if (v == p) continue;
    long long int part = dfs1(v, nd);
    sum += part;
    total += part * (n - part);
  }
  return sum;
}
void dfs2(int nd, int p) {
  for (int i = 0; i < (int)tree[nd].size(); i++) {
    int v = tree[nd][i];
    if (v == p) continue;
    dfs2(v, nd);
    for (int x = 0; x < k; x++) {
      for (int y = 1; y <= k; y++) {
        int ym = (y % k);
        ym = (ym ? ym : k);
        int yy = ym - 1;
        int nm = (x + ym) % k;
        long long int a = dp[nd][x];
        long long int b = dp[v][yy];
        long long int c = (ym == (1 % k));
        modSum[nm] += (a * (b + c));
      }
    }
    dp[nd][1 % k]++;
    modSum[1 % k]++;
    for (int j = 0; j < k - 1; j++) {
      dp[nd][j + 1] += dp[v][j];
      modSum[j + 1] += dp[v][j];
    }
    dp[nd][0] += dp[v][k - 1];
    modSum[0] += dp[v][k - 1];
  }
}
int main() {
  scanf("%d %d", &n, &k);
  int a, b;
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &a, &b);
    tree[a].push_back(b);
    tree[b].push_back(a);
  }
  memset(modSum, 0, sizeof(modSum));
  memset(dp, 0, sizeof(dp));
  dfs1(1, 0);
  dfs2(1, 0);
  for (int i = 1; i < k; i++) {
    total += (modSum[i] * (k - i));
  }
  printf("%I64d", total / k);
  return 0;
}

```