#include <bits/stdc++.h>
using namespace std;
vector<int> v[300000];
long long ans = 0;
int n, k;
pair<long long, vector<long long> > num[300000];
void add(pair<long long, vector<long long> >& a,
         pair<long long, vector<long long> > b) {
  a.first += b.first;
  for (int i = 0; i < k; ++i) a.second[i] += b.second[i];
}
void sub(pair<long long, vector<long long> >& a,
         pair<long long, vector<long long> > b) {
  a.first -= b.first;
  for (int i = 0; i < k; ++i) a.second[i] -= b.second[i];
}
pair<long long, vector<long long> > step(
    pair<long long, vector<long long> > a) {
  a.second.back() += 1;
  a.first += a.second.back();
  a.second.push_back(a.second.back());
  for (int i = k - 2; i >= 0; --i) a.second[i + 1] = a.second[i];
  a.second[0] = a.second.back();
  a.second.pop_back();
  return a;
}
pair<long long, vector<long long> > dfs(int x, int last = -1) {
  num[x] = make_pair(0LL, vector<long long>(k, 0));
  for (auto ed : v[x])
    if (ed != last) {
      add(num[x], step(dfs(ed, x)));
    }
  return num[x];
}
void print(pair<long long, vector<long long> > up) {}
void calc(int x, pair<long long, vector<long long> > up, int last = -1) {
  if (last != -1) up = step(up);
  for (auto ed : v[x])
    if (ed != last) add(up, step(num[ed]));
  ans += up.first;
  for (auto ed : v[x])
    if (ed != last) {
      sub(up, step(num[ed]));
      calc(ed, up, x);
      add(up, step(num[ed]));
    }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    int fr, to;
    cin >> fr >> to;
    --fr, --to;
    v[fr].push_back(to);
    v[to].push_back(fr);
  }
  dfs(0);
  calc(0, make_pair(0LL, vector<long long>(k, 0)));
  cout << ans / 2 << "\n";
}

### Prompt

Please create a solution in Cpp 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;
vector<int> v[300000];
long long ans = 0;
int n, k;
pair<long long, vector<long long> > num[300000];
void add(pair<long long, vector<long long> >& a,
         pair<long long, vector<long long> > b) {
  a.first += b.first;
  for (int i = 0; i < k; ++i) a.second[i] += b.second[i];
}
void sub(pair<long long, vector<long long> >& a,
         pair<long long, vector<long long> > b) {
  a.first -= b.first;
  for (int i = 0; i < k; ++i) a.second[i] -= b.second[i];
}
pair<long long, vector<long long> > step(
    pair<long long, vector<long long> > a) {
  a.second.back() += 1;
  a.first += a.second.back();
  a.second.push_back(a.second.back());
  for (int i = k - 2; i >= 0; --i) a.second[i + 1] = a.second[i];
  a.second[0] = a.second.back();
  a.second.pop_back();
  return a;
}
pair<long long, vector<long long> > dfs(int x, int last = -1) {
  num[x] = make_pair(0LL, vector<long long>(k, 0));
  for (auto ed : v[x])
    if (ed != last) {
      add(num[x], step(dfs(ed, x)));
    }
  return num[x];
}
void print(pair<long long, vector<long long> > up) {}
void calc(int x, pair<long long, vector<long long> > up, int last = -1) {
  if (last != -1) up = step(up);
  for (auto ed : v[x])
    if (ed != last) add(up, step(num[ed]));
  ans += up.first;
  for (auto ed : v[x])
    if (ed != last) {
      sub(up, step(num[ed]));
      calc(ed, up, x);
      add(up, step(num[ed]));
    }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    int fr, to;
    cin >> fr >> to;
    --fr, --to;
    v[fr].push_back(to);
    v[to].push_back(fr);
  }
  dfs(0);
  calc(0, make_pair(0LL, vector<long long>(k, 0)));
  cout << ans / 2 << "\n";
}

```