#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using pll = pair<ll, ll>;
using vpll = vector<pll>;
static bool constexpr dbg = 0;
ll constexpr NM = 2.1e5;
ll n, k;
vector<ll> adj[NM];
ll cnt[NM];
ll mc[NM][5];
ll sum;
ll constexpr root = 1;
void dfs1(ll c, ll p) {
  cnt[c] = 1;
  mc[c][0] = 1;
  for (auto a : adj[c])
    if (a != p) {
      dfs1(a, c);
      cnt[c] += cnt[a];
      for (ll i = 0; i < (ll)(k); ++i) mc[c][(i + 1) % k] += mc[a][i];
    }
  sum += 2 * cnt[c] * (n - cnt[c]);
}
void dfs2(ll c, ll p) {
  ll pa[5] = {0};
  if (c != root) {
    for (ll i = 0; i < (ll)(k); ++i) pa[i] = mc[p][i] - mc[c][(i - 1 + k) % k];
    for (ll i = 0; i < (ll)(k); ++i)
      if ((i + 1) % k) sum += (k - i - 1) * (pa[i] + mc[c][i]);
  }
  for (ll i = 0; i < (ll)(k); ++i) mc[c][(i + 1) % k] += pa[i];
  for (auto a : adj[c])
    if (a != p) {
      dfs2(a, c);
    }
}
int main() {
  if (!dbg) {
    std::ios::sync_with_stdio(false);
    cin.tie(nullptr);
  }
  cin >> n >> k;
  for (ll _ = 1; _ <= (ll)(n - 1); ++_) {
    ll x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  cout << sum / k / 2 << endl;
  return 0;
}

### 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;
using ll = long long;
using ull = unsigned long long;
using pll = pair<ll, ll>;
using vpll = vector<pll>;
static bool constexpr dbg = 0;
ll constexpr NM = 2.1e5;
ll n, k;
vector<ll> adj[NM];
ll cnt[NM];
ll mc[NM][5];
ll sum;
ll constexpr root = 1;
void dfs1(ll c, ll p) {
  cnt[c] = 1;
  mc[c][0] = 1;
  for (auto a : adj[c])
    if (a != p) {
      dfs1(a, c);
      cnt[c] += cnt[a];
      for (ll i = 0; i < (ll)(k); ++i) mc[c][(i + 1) % k] += mc[a][i];
    }
  sum += 2 * cnt[c] * (n - cnt[c]);
}
void dfs2(ll c, ll p) {
  ll pa[5] = {0};
  if (c != root) {
    for (ll i = 0; i < (ll)(k); ++i) pa[i] = mc[p][i] - mc[c][(i - 1 + k) % k];
    for (ll i = 0; i < (ll)(k); ++i)
      if ((i + 1) % k) sum += (k - i - 1) * (pa[i] + mc[c][i]);
  }
  for (ll i = 0; i < (ll)(k); ++i) mc[c][(i + 1) % k] += pa[i];
  for (auto a : adj[c])
    if (a != p) {
      dfs2(a, c);
    }
}
int main() {
  if (!dbg) {
    std::ios::sync_with_stdio(false);
    cin.tie(nullptr);
  }
  cin >> n >> k;
  for (ll _ = 1; _ <= (ll)(n - 1); ++_) {
    ll x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  cout << sum / k / 2 << endl;
  return 0;
}

```