#include <bits/stdc++.h>
using namespace std;
const int maxn = 234567, maxk = 5;
int n, k;
long long subtreeSize[maxn], totalDis[maxn];
vector<long long> e[maxn], numSubtreeLenModK[maxn], numLenModK[maxn];
void getSize(int id, int p) {
  subtreeSize[id] = 1;
  for (int i = 0; i < e[id].size(); i++)
    if (e[id][i] != p) {
      getSize(e[id][i], id);
      subtreeSize[id] += subtreeSize[e[id][i]];
    }
}
void getDisToRoot(int id, int p, int d) {
  totalDis[1] += d;
  for (int i = 0; i < e[id].size(); i++)
    if (e[id][i] != p) getDisToRoot(e[id][i], id, d + 1);
}
void getTotalDis(int id, int p) {
  for (int i = 0; i < e[id].size(); i++)
    if (e[id][i] != p) {
      totalDis[e[id][i]] = totalDis[id] - subtreeSize[e[id][i]] * 2 + n;
      getTotalDis(e[id][i], id);
    }
}
void getNumSubtreeLenModK(int id, int p) {
  numSubtreeLenModK[id] = vector<long long>(k, 0);
  numSubtreeLenModK[id][0] = 1;
  for (int i = 0; i < e[id].size(); i++)
    if (e[id][i] != p) {
      getNumSubtreeLenModK(e[id][i], id);
      for (int j = 0; j < k; j++)
        numSubtreeLenModK[id][j] +=
            numSubtreeLenModK[e[id][i]][(j + k - 1) % k];
    }
}
void getNumLenModK(int id, int p, int d) {
  numLenModK[1][d % k]++;
  for (int i = 0; i < e[id].size(); i++)
    if (e[id][i] != p) getNumLenModK(e[id][i], id, d + 1);
}
void getTotalNumLenModK(int id, int p) {
  for (int i = 0; i < e[id].size(); i++)
    if (e[id][i] != p) {
      numLenModK[e[id][i]] = vector<long long>(k, 0);
      auto tmp = numLenModK[id];
      for (int j = 0; j < k; j++)
        tmp[j] -= numSubtreeLenModK[e[id][i]][(j + k - 1) % k];
      for (int j = 0; j < k; j++)
        numLenModK[e[id][i]][j] += tmp[(j + k - 1) % k];
      for (int j = 0; j < k; j++)
        numLenModK[e[id][i]][j] += numSubtreeLenModK[e[id][i]][j];
      getTotalNumLenModK(e[id][i], id);
    }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    e[a].push_back(b);
    e[b].push_back(a);
  }
  getSize(1, 1);
  getDisToRoot(1, 1, 0);
  getTotalDis(1, 1);
  getNumSubtreeLenModK(1, 1);
  numLenModK[1] = vector<long long>(k, 0);
  getNumLenModK(1, 1, 0);
  getTotalNumLenModK(1, 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long tmp = totalDis[i];
    for (int j = 1; j < k; j++) tmp += numLenModK[i][j] * (k - j);
    ans += tmp / k;
  }
  cout << ans / 2 << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 maxn = 234567, maxk = 5;
int n, k;
long long subtreeSize[maxn], totalDis[maxn];
vector<long long> e[maxn], numSubtreeLenModK[maxn], numLenModK[maxn];
void getSize(int id, int p) {
  subtreeSize[id] = 1;
  for (int i = 0; i < e[id].size(); i++)
    if (e[id][i] != p) {
      getSize(e[id][i], id);
      subtreeSize[id] += subtreeSize[e[id][i]];
    }
}
void getDisToRoot(int id, int p, int d) {
  totalDis[1] += d;
  for (int i = 0; i < e[id].size(); i++)
    if (e[id][i] != p) getDisToRoot(e[id][i], id, d + 1);
}
void getTotalDis(int id, int p) {
  for (int i = 0; i < e[id].size(); i++)
    if (e[id][i] != p) {
      totalDis[e[id][i]] = totalDis[id] - subtreeSize[e[id][i]] * 2 + n;
      getTotalDis(e[id][i], id);
    }
}
void getNumSubtreeLenModK(int id, int p) {
  numSubtreeLenModK[id] = vector<long long>(k, 0);
  numSubtreeLenModK[id][0] = 1;
  for (int i = 0; i < e[id].size(); i++)
    if (e[id][i] != p) {
      getNumSubtreeLenModK(e[id][i], id);
      for (int j = 0; j < k; j++)
        numSubtreeLenModK[id][j] +=
            numSubtreeLenModK[e[id][i]][(j + k - 1) % k];
    }
}
void getNumLenModK(int id, int p, int d) {
  numLenModK[1][d % k]++;
  for (int i = 0; i < e[id].size(); i++)
    if (e[id][i] != p) getNumLenModK(e[id][i], id, d + 1);
}
void getTotalNumLenModK(int id, int p) {
  for (int i = 0; i < e[id].size(); i++)
    if (e[id][i] != p) {
      numLenModK[e[id][i]] = vector<long long>(k, 0);
      auto tmp = numLenModK[id];
      for (int j = 0; j < k; j++)
        tmp[j] -= numSubtreeLenModK[e[id][i]][(j + k - 1) % k];
      for (int j = 0; j < k; j++)
        numLenModK[e[id][i]][j] += tmp[(j + k - 1) % k];
      for (int j = 0; j < k; j++)
        numLenModK[e[id][i]][j] += numSubtreeLenModK[e[id][i]][j];
      getTotalNumLenModK(e[id][i], id);
    }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    e[a].push_back(b);
    e[b].push_back(a);
  }
  getSize(1, 1);
  getDisToRoot(1, 1, 0);
  getTotalDis(1, 1);
  getNumSubtreeLenModK(1, 1);
  numLenModK[1] = vector<long long>(k, 0);
  getNumLenModK(1, 1, 0);
  getTotalNumLenModK(1, 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long tmp = totalDis[i];
    for (int j = 1; j < k; j++) tmp += numLenModK[i][j] * (k - j);
    ans += tmp / k;
  }
  cout << ans / 2 << endl;
  return 0;
}

```