#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
long long n, k;
long long ans = 0;
vector<vector<long long> > g;
pair<long long, vector<long long> > dfs(long long v, long long pr,
                                        long long h = 0) {
  vector<long long> d(k, 0);
  long long sz = 1;
  d[h % k] = 1;
  for (long long i = 0; i < g[v].size(); i++) {
    if (g[v][i] == pr) continue;
    pair<long long, vector<long long> > kek = dfs(g[v][i], v, h + 1);
    sz += kek.first;
    ans += kek.first * (n - kek.first);
    for (long long j = 0; j < k; j++) {
      for (long long y = 0; y < k; y++) {
        ans += ((k - ((y + j - ((2 * h) % k) + k) % k) + k) % k) * d[j] *
               kek.second[y];
      }
    }
    for (long long j = 0; j < k; j++) d[j] += kek.second[j];
  }
  return {sz, d};
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  g.resize(n);
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, 0);
  cout << ans / k;
  return 0;
}

### Prompt

Please formulate a CPP solution 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;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
long long n, k;
long long ans = 0;
vector<vector<long long> > g;
pair<long long, vector<long long> > dfs(long long v, long long pr,
                                        long long h = 0) {
  vector<long long> d(k, 0);
  long long sz = 1;
  d[h % k] = 1;
  for (long long i = 0; i < g[v].size(); i++) {
    if (g[v][i] == pr) continue;
    pair<long long, vector<long long> > kek = dfs(g[v][i], v, h + 1);
    sz += kek.first;
    ans += kek.first * (n - kek.first);
    for (long long j = 0; j < k; j++) {
      for (long long y = 0; y < k; y++) {
        ans += ((k - ((y + j - ((2 * h) % k) + k) % k) + k) % k) * d[j] *
               kek.second[y];
      }
    }
    for (long long j = 0; j < k; j++) d[j] += kek.second[j];
  }
  return {sz, d};
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  g.resize(n);
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, 0);
  cout << ans / k;
  return 0;
}

```