#include <bits/stdc++.h>
using namespace std;
using f80 = long double;
using f64 = double;
using i64 = long long;
using i64 = long long;
const int N = 2e5 + 5, K = 6;
int n, k, t, ctr, siz;
i64 sum;
vector<int> g[N], csub[N];
bool tkn[N];
int dst[N], sub[N];
int getsiz(int u, int far = 0) {
  int w = 1;
  for (auto v : g[u])
    if (v != far && !tkn[v]) w += getsiz(v, u);
  return w;
}
void getdst(int u, int far = 0, int d = 0) {
  dst[u] = d;
  for (auto v : g[u])
    if (v != far && !tkn[v]) getdst(v, u, d + 1);
}
void getsub(int u, int far = 0) {
  csub[t].push_back(u);
  for (auto v : g[u])
    if (v != far && !tkn[v]) getsub(v, u);
}
void centroid(int u, int far = 0) {
  sub[u] = 1;
  for (auto v : g[u])
    if (v != far && !tkn[v]) {
      centroid(v, u);
      sub[u] += sub[v];
    }
  if ((siz - sub[u]) > siz / 2 + 1) return;
  for (auto v : g[u])
    if (v != far && !tkn[v])
      if (sub[v] > siz / 2 + 1) return;
  ctr = u;
}
i64 square(const vector<int> &v) {
  vector<int> f(k, 0);
  i64 t = 0, s = 0;
  for (auto i : v) {
    s += dst[i] / k;
    ++f[dst[i] % k];
  }
  s *= 2 * v.size();
  for (int i = 0; i < k; ++i)
    for (int j = 0; j < k; ++j) s += 1LL * f[i] * f[j] * ((i + j + k - 1) / k);
  return s;
}
void divimp(int u, int far = 0) {
  map<int, int> mp;
  csub[u].clear();
  csub[u].push_back(u);
  tkn[u] = 1;
  dst[u] = 0;
  getdst(t = u);
  for (auto v : g[u])
    if (!tkn[v]) {
      siz = getsiz(v);
      centroid(v);
      csub[v].clear();
      getsub(t = v);
      mp[v] = ctr;
      csub[u].insert(end(csub[u]), begin(csub[t]), end(csub[t]));
    }
  sum += square(csub[u]);
  for (auto v : g[u])
    if (v != far && !tkn[v]) sum -= square(csub[v]);
  for (auto v : g[u])
    if (v != far && !tkn[v]) divimp(mp[v], u);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int a, b;
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  divimp(1);
  cout << sum / 2 << '\n';
  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;
using f80 = long double;
using f64 = double;
using i64 = long long;
using i64 = long long;
const int N = 2e5 + 5, K = 6;
int n, k, t, ctr, siz;
i64 sum;
vector<int> g[N], csub[N];
bool tkn[N];
int dst[N], sub[N];
int getsiz(int u, int far = 0) {
  int w = 1;
  for (auto v : g[u])
    if (v != far && !tkn[v]) w += getsiz(v, u);
  return w;
}
void getdst(int u, int far = 0, int d = 0) {
  dst[u] = d;
  for (auto v : g[u])
    if (v != far && !tkn[v]) getdst(v, u, d + 1);
}
void getsub(int u, int far = 0) {
  csub[t].push_back(u);
  for (auto v : g[u])
    if (v != far && !tkn[v]) getsub(v, u);
}
void centroid(int u, int far = 0) {
  sub[u] = 1;
  for (auto v : g[u])
    if (v != far && !tkn[v]) {
      centroid(v, u);
      sub[u] += sub[v];
    }
  if ((siz - sub[u]) > siz / 2 + 1) return;
  for (auto v : g[u])
    if (v != far && !tkn[v])
      if (sub[v] > siz / 2 + 1) return;
  ctr = u;
}
i64 square(const vector<int> &v) {
  vector<int> f(k, 0);
  i64 t = 0, s = 0;
  for (auto i : v) {
    s += dst[i] / k;
    ++f[dst[i] % k];
  }
  s *= 2 * v.size();
  for (int i = 0; i < k; ++i)
    for (int j = 0; j < k; ++j) s += 1LL * f[i] * f[j] * ((i + j + k - 1) / k);
  return s;
}
void divimp(int u, int far = 0) {
  map<int, int> mp;
  csub[u].clear();
  csub[u].push_back(u);
  tkn[u] = 1;
  dst[u] = 0;
  getdst(t = u);
  for (auto v : g[u])
    if (!tkn[v]) {
      siz = getsiz(v);
      centroid(v);
      csub[v].clear();
      getsub(t = v);
      mp[v] = ctr;
      csub[u].insert(end(csub[u]), begin(csub[t]), end(csub[t]));
    }
  sum += square(csub[u]);
  for (auto v : g[u])
    if (v != far && !tkn[v]) sum -= square(csub[v]);
  for (auto v : g[u])
    if (v != far && !tkn[v]) divimp(mp[v], u);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int a, b;
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  divimp(1);
  cout << sum / 2 << '\n';
  return 0;
}

```