#include <bits/stdc++.h>
using namespace std;
const int MN = 200001;
const int MK = 6;
int K;
struct edge {
  int to, p_id;
  edge(int a, int b) : to(a), p_id(b) {}
};
struct state {
  bool seen;
  long long missing;
  long long total;
  vector<long long> partial;
  state() { clear(); }
  void clear() {
    seen = false;
    missing = 0;
    total = 0;
    partial.clear();
  }
};
state dp1[MN];
state dp2[MK][MN];
vector<edge> g[MN];
void add_edge(int u, int v) {
  int id_u_v = g[u].size();
  int id_v_u = g[v].size();
  g[u].emplace_back(v, id_v_u);
  g[v].emplace_back(u, id_u_v);
}
int go_size(int node, int id_parent) {
  state &s = dp1[node];
  if (!s.seen) {
    int ans = 1;
    s.partial.assign(g[node].size(), 0);
    for (int i = 0; i < int(g[node].size()); i++) {
      int to = g[node][i].to;
      int pid = g[node][i].p_id;
      if (i != id_parent) {
        int tmp = go_size(to, pid);
        ans += tmp;
        s.partial[i] = tmp;
      }
    }
    s.missing = id_parent;
    s.total = ans;
    s.seen = true;
    return ans;
  } else {
    if (s.missing == id_parent) {
      return s.total;
    }
    if (s.missing != -1) {
      int tmp = go_size(g[node][s.missing].to, g[node][s.missing].p_id);
      s.partial[s.missing] = tmp;
      s.total += tmp;
      s.missing = -1;
    }
    int extra = (id_parent == -1) ? 0 : s.partial[id_parent];
    return s.total - extra;
  }
}
long long go(int node, int id_parent, int x) {
  state &s = dp2[x][node];
  if (!s.seen) {
    long long ans = 0;
    s.partial.assign(g[node].size(), 0);
    for (int i = 0; i < int(g[node].size()); i++) {
      int to = g[node][i].to;
      int pid = g[node][i].p_id;
      if (i != id_parent) {
        long long tmp = 0;
        if (x)
          tmp = go(to, pid, x - 1);
        else
          tmp = go(to, pid, K - 1) + go_size(to, pid);
        ans += tmp;
        s.partial[i] = tmp;
      }
    }
    s.missing = id_parent;
    s.total = ans;
    s.seen = true;
    return ans;
  } else {
    if (s.missing == id_parent) {
      return s.total;
    }
    if (s.missing != -1) {
      long long tmp = 0;
      int to = g[node][s.missing].to, pid = g[node][s.missing].p_id;
      if (x)
        tmp = go(to, pid, x - 1);
      else
        tmp = go(to, pid, K - 1) + go_size(to, pid);
      s.partial[s.missing] = tmp;
      s.total += tmp;
      s.missing = -1;
    }
    long long extra = (id_parent == -1) ? 0 : s.partial[id_parent];
    return s.total - extra;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n >> K;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    add_edge(u, v);
  }
  for (int i = 0; i < n; i++) {
    dp1[i].clear();
    for (int j = 0; j < MK; j++) dp2[j][i].clear();
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) ans += go(i, -1, 0);
  cout << (ans >> 1) << '\n';
  return 0;
}

### Prompt

Develop a solution in CPP 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 MN = 200001;
const int MK = 6;
int K;
struct edge {
  int to, p_id;
  edge(int a, int b) : to(a), p_id(b) {}
};
struct state {
  bool seen;
  long long missing;
  long long total;
  vector<long long> partial;
  state() { clear(); }
  void clear() {
    seen = false;
    missing = 0;
    total = 0;
    partial.clear();
  }
};
state dp1[MN];
state dp2[MK][MN];
vector<edge> g[MN];
void add_edge(int u, int v) {
  int id_u_v = g[u].size();
  int id_v_u = g[v].size();
  g[u].emplace_back(v, id_v_u);
  g[v].emplace_back(u, id_u_v);
}
int go_size(int node, int id_parent) {
  state &s = dp1[node];
  if (!s.seen) {
    int ans = 1;
    s.partial.assign(g[node].size(), 0);
    for (int i = 0; i < int(g[node].size()); i++) {
      int to = g[node][i].to;
      int pid = g[node][i].p_id;
      if (i != id_parent) {
        int tmp = go_size(to, pid);
        ans += tmp;
        s.partial[i] = tmp;
      }
    }
    s.missing = id_parent;
    s.total = ans;
    s.seen = true;
    return ans;
  } else {
    if (s.missing == id_parent) {
      return s.total;
    }
    if (s.missing != -1) {
      int tmp = go_size(g[node][s.missing].to, g[node][s.missing].p_id);
      s.partial[s.missing] = tmp;
      s.total += tmp;
      s.missing = -1;
    }
    int extra = (id_parent == -1) ? 0 : s.partial[id_parent];
    return s.total - extra;
  }
}
long long go(int node, int id_parent, int x) {
  state &s = dp2[x][node];
  if (!s.seen) {
    long long ans = 0;
    s.partial.assign(g[node].size(), 0);
    for (int i = 0; i < int(g[node].size()); i++) {
      int to = g[node][i].to;
      int pid = g[node][i].p_id;
      if (i != id_parent) {
        long long tmp = 0;
        if (x)
          tmp = go(to, pid, x - 1);
        else
          tmp = go(to, pid, K - 1) + go_size(to, pid);
        ans += tmp;
        s.partial[i] = tmp;
      }
    }
    s.missing = id_parent;
    s.total = ans;
    s.seen = true;
    return ans;
  } else {
    if (s.missing == id_parent) {
      return s.total;
    }
    if (s.missing != -1) {
      long long tmp = 0;
      int to = g[node][s.missing].to, pid = g[node][s.missing].p_id;
      if (x)
        tmp = go(to, pid, x - 1);
      else
        tmp = go(to, pid, K - 1) + go_size(to, pid);
      s.partial[s.missing] = tmp;
      s.total += tmp;
      s.missing = -1;
    }
    long long extra = (id_parent == -1) ? 0 : s.partial[id_parent];
    return s.total - extra;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n >> K;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    add_edge(u, v);
  }
  for (int i = 0; i < n; i++) {
    dp1[i].clear();
    for (int j = 0; j < MK; j++) dp2[j][i].clear();
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) ans += go(i, -1, 0);
  cout << (ans >> 1) << '\n';
  return 0;
}

```