#include <bits/stdc++.h>
using namespace std;
const int OO = 1e9;
const double EPS = 1e-9;
template <class T>
void _db(const char* dbStr, T e) {
  cout << dbStr << " = " << e << endl;
}
template <class T, class... L>
void _db(const char* dbStr, T e, L... r) {
  while (*dbStr != ',') cout << *dbStr++;
  cout << " = " << e << ',';
  _db(dbStr + 1, r...);
}
template <class S, class T>
ostream& operator<<(ostream& o, const map<S, T>& v) {
  o << "[";
  int i = 0;
  for (const pair<S, T>& pr : v)
    o << (!i++ ? "" : ", ") << "{" << pr.first << " : " << pr.second << "}";
  return o << "]";
}
template <template <class, class...> class S, class T, class... L>
ostream& operator<<(ostream& o, const S<T, L...>& v) {
  o << "[";
  int i = 0;
  for (const auto& e : v) o << (!i++ ? "" : ", ") << e;
  return o << "]";
}
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& pr) {
  return o << "(" << pr.first << ", " << pr.second << ")";
}
ostream& operator<<(ostream& o, const string& s) {
  for (const char& c : s) o << c;
  return o;
}
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
template <class T>
using VVV = VV<V<T>>;
using ll = long long;
using pii = pair<int, int>;
using vi = V<int>;
using vii = V<pii>;
using vvi = VV<int>;
using mii = map<int, int>;
using umii = unordered_map<int, int>;
using si = set<int>;
using usi = unordered_set<int>;
const int MX = 2e5 + 5;
int n, k, subTr[MX][5];
ll sumD0, ans;
vvi adj;
void dfs0(int u, int p, int d) {
  subTr[u][0]++;
  sumD0 += (d + k - 1) / k;
  for (auto& v : adj[u]) {
    if (v != p) {
      dfs0(v, u, d + 1);
      for (int i = 0; i < k; ++i) {
        subTr[u][(i + 1) % k] += subTr[v][i];
      }
    }
  }
}
void dfs(int u, int p, ll sumD, vi distMod) {
  true;
  ans += sumD;
  for (auto& v : adj[u]) {
    if (v != p) {
      sumD += distMod[0] - subTr[v][k - 1];
      sumD -= subTr[v][0];
      vi nextDistMod(k);
      for (int i = 0; i < k; ++i) {
        nextDistMod[(i + 1) % k] += distMod[i] - subTr[v][(i - 1 + k) % k];
        nextDistMod[i] += subTr[v][i];
      }
      true;
      dfs(v, u, sumD, nextDistMod);
      sumD -= distMod[0] - subTr[v][k - 1];
      sumD += subTr[v][0];
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  cin.tie(0);
  cin >> n >> k;
  adj.assign(n + 1, vi());
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs0(1, -1, 0);
  vi distMod;
  for (int i = 0; i < k; ++i) {
    distMod.push_back(subTr[1][i]);
  }
  dfs(1, -1, sumD0, distMod);
  true;
  cout << ans / 2 << '\n';
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int OO = 1e9;
const double EPS = 1e-9;
template <class T>
void _db(const char* dbStr, T e) {
  cout << dbStr << " = " << e << endl;
}
template <class T, class... L>
void _db(const char* dbStr, T e, L... r) {
  while (*dbStr != ',') cout << *dbStr++;
  cout << " = " << e << ',';
  _db(dbStr + 1, r...);
}
template <class S, class T>
ostream& operator<<(ostream& o, const map<S, T>& v) {
  o << "[";
  int i = 0;
  for (const pair<S, T>& pr : v)
    o << (!i++ ? "" : ", ") << "{" << pr.first << " : " << pr.second << "}";
  return o << "]";
}
template <template <class, class...> class S, class T, class... L>
ostream& operator<<(ostream& o, const S<T, L...>& v) {
  o << "[";
  int i = 0;
  for (const auto& e : v) o << (!i++ ? "" : ", ") << e;
  return o << "]";
}
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& pr) {
  return o << "(" << pr.first << ", " << pr.second << ")";
}
ostream& operator<<(ostream& o, const string& s) {
  for (const char& c : s) o << c;
  return o;
}
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
template <class T>
using VVV = VV<V<T>>;
using ll = long long;
using pii = pair<int, int>;
using vi = V<int>;
using vii = V<pii>;
using vvi = VV<int>;
using mii = map<int, int>;
using umii = unordered_map<int, int>;
using si = set<int>;
using usi = unordered_set<int>;
const int MX = 2e5 + 5;
int n, k, subTr[MX][5];
ll sumD0, ans;
vvi adj;
void dfs0(int u, int p, int d) {
  subTr[u][0]++;
  sumD0 += (d + k - 1) / k;
  for (auto& v : adj[u]) {
    if (v != p) {
      dfs0(v, u, d + 1);
      for (int i = 0; i < k; ++i) {
        subTr[u][(i + 1) % k] += subTr[v][i];
      }
    }
  }
}
void dfs(int u, int p, ll sumD, vi distMod) {
  true;
  ans += sumD;
  for (auto& v : adj[u]) {
    if (v != p) {
      sumD += distMod[0] - subTr[v][k - 1];
      sumD -= subTr[v][0];
      vi nextDistMod(k);
      for (int i = 0; i < k; ++i) {
        nextDistMod[(i + 1) % k] += distMod[i] - subTr[v][(i - 1 + k) % k];
        nextDistMod[i] += subTr[v][i];
      }
      true;
      dfs(v, u, sumD, nextDistMod);
      sumD -= distMod[0] - subTr[v][k - 1];
      sumD += subTr[v][0];
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  cin.tie(0);
  cin >> n >> k;
  adj.assign(n + 1, vi());
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs0(1, -1, 0);
  vi distMod;
  for (int i = 0; i < k; ++i) {
    distMod.push_back(subTr[1][i]);
  }
  dfs(1, -1, sumD0, distMod);
  true;
  cout << ans / 2 << '\n';
  return 0;
}

```