#include <bits/stdc++.h>
using namespace std;
vector<int> adia[200010];
vector<int> suma[200010];
long long pasi[200010];
long long ans;
int n, k;
void dfs1(int nod, int tata);
void dfs2(int nod, int tata, long long pasi_act, vector<int> pasi_tata);
void add(vector<int>& x, vector<int>& y, long long& pasi);
void scoate(vector<int>& x, vector<int>& y, long long& pasi);
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  int a, b;
  while (--n) {
    cin >> a >> b;
    adia[a].push_back(b);
    adia[b].push_back(a);
  }
  dfs1(1, 0);
  vector<int> v(k + 1, 0);
  dfs2(1, 0, 0, v);
  cout << ans / 2ll;
  return 0;
}
void dfs2(int nod, int tata, long long pasi_act, vector<int> pasi_tata) {
  vector<int> _pasi = pasi_tata;
  for (int i(1); i < k; i++) pasi_tata[i + 1] = _pasi[i];
  pasi_tata[1] = _pasi[k];
  pasi_act += _pasi[k];
  for (auto i : adia[nod]) {
    if (i != tata) {
      add(suma[i], pasi_tata, pasi_act);
      pasi_act += pasi[i];
    }
  }
  ans += pasi_act;
  pasi_tata[k]++;
  for (auto i : adia[nod]) {
    if (i != tata) {
      scoate(suma[i], pasi_tata, pasi_act);
      pasi_act -= pasi[i];
      dfs2(i, nod, pasi_act, pasi_tata);
      add(suma[i], pasi_tata, pasi_act);
      pasi_act += pasi[i];
    }
  }
}
void dfs1(int nod, int tata) {
  suma[nod] = vector<int>(k + 1, 0);
  for (auto i : adia[nod]) {
    if (i != tata) {
      dfs1(i, nod);
      add(suma[i], suma[nod], pasi[nod]);
      pasi[nod] += pasi[i];
    }
  }
}
void scoate(vector<int>& x, vector<int>& y, long long& pasi) {
  y[1]--;
  pasi -= 1ll * x[k] + 1ll;
  for (int i(1); i < k; i++) y[i + 1] -= x[i];
  y[1] -= x[k];
}
void add(vector<int>& x, vector<int>& y, long long& pasi) {
  pasi += 1ll * x[k] + 1ll;
  for (int i(1); i < k; i++) y[i + 1] += x[i];
  y[1] += x[k] + 1ll;
}

### 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> adia[200010];
vector<int> suma[200010];
long long pasi[200010];
long long ans;
int n, k;
void dfs1(int nod, int tata);
void dfs2(int nod, int tata, long long pasi_act, vector<int> pasi_tata);
void add(vector<int>& x, vector<int>& y, long long& pasi);
void scoate(vector<int>& x, vector<int>& y, long long& pasi);
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  int a, b;
  while (--n) {
    cin >> a >> b;
    adia[a].push_back(b);
    adia[b].push_back(a);
  }
  dfs1(1, 0);
  vector<int> v(k + 1, 0);
  dfs2(1, 0, 0, v);
  cout << ans / 2ll;
  return 0;
}
void dfs2(int nod, int tata, long long pasi_act, vector<int> pasi_tata) {
  vector<int> _pasi = pasi_tata;
  for (int i(1); i < k; i++) pasi_tata[i + 1] = _pasi[i];
  pasi_tata[1] = _pasi[k];
  pasi_act += _pasi[k];
  for (auto i : adia[nod]) {
    if (i != tata) {
      add(suma[i], pasi_tata, pasi_act);
      pasi_act += pasi[i];
    }
  }
  ans += pasi_act;
  pasi_tata[k]++;
  for (auto i : adia[nod]) {
    if (i != tata) {
      scoate(suma[i], pasi_tata, pasi_act);
      pasi_act -= pasi[i];
      dfs2(i, nod, pasi_act, pasi_tata);
      add(suma[i], pasi_tata, pasi_act);
      pasi_act += pasi[i];
    }
  }
}
void dfs1(int nod, int tata) {
  suma[nod] = vector<int>(k + 1, 0);
  for (auto i : adia[nod]) {
    if (i != tata) {
      dfs1(i, nod);
      add(suma[i], suma[nod], pasi[nod]);
      pasi[nod] += pasi[i];
    }
  }
}
void scoate(vector<int>& x, vector<int>& y, long long& pasi) {
  y[1]--;
  pasi -= 1ll * x[k] + 1ll;
  for (int i(1); i < k; i++) y[i + 1] -= x[i];
  y[1] -= x[k];
}
void add(vector<int>& x, vector<int>& y, long long& pasi) {
  pasi += 1ll * x[k] + 1ll;
  for (int i(1); i < k; i++) y[i + 1] += x[i];
  y[1] += x[k] + 1ll;
}

```