#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 2e5 + 5;
struct P {
  long long n, s;
  P() {}
  P(long long n, long long s) : n(n), s(s) {}
};
long long n, k;
vector<long long> g[MAXN];
long long ans;
vector<P> dfs(long long u, long long p) {
  vector<P> res(k, P(0, 0));
  for (long long v : g[u]) {
    if (v == p) continue;
    vector<P> vec1 = dfs(v, u);
    vector<P> vec2(k);
    vec2[1 % k] = P(vec1[0].n, vec1[0].s + vec1[0].n);
    for (long long i = 1; i < k; i++)
      vec2[(i + 1) % k] = P(vec1[i].n, vec1[i].s);
    for (long long i = 0; i < k; i++)
      for (long long j = 0; j < k; j++)
        ans += res[i].s * vec2[j].n + res[i].n * vec2[j].s -
               res[i].n * vec2[j].n * ((i != 0) && (j != 0) && (i + j <= k));
    for (long long i = 0; i < k; i++)
      res[i] = P(res[i].n + vec2[i].n, res[i].s + vec2[i].s);
  }
  for (long long i = 0; i < k; i++) ans += res[i].s;
  res[0] = P(res[0].n + 1, res[0].s);
  return res;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (long long i = 0; i < n - 1; i++) {
    static long long u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, -1);
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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;
const long long MAXN = 2e5 + 5;
struct P {
  long long n, s;
  P() {}
  P(long long n, long long s) : n(n), s(s) {}
};
long long n, k;
vector<long long> g[MAXN];
long long ans;
vector<P> dfs(long long u, long long p) {
  vector<P> res(k, P(0, 0));
  for (long long v : g[u]) {
    if (v == p) continue;
    vector<P> vec1 = dfs(v, u);
    vector<P> vec2(k);
    vec2[1 % k] = P(vec1[0].n, vec1[0].s + vec1[0].n);
    for (long long i = 1; i < k; i++)
      vec2[(i + 1) % k] = P(vec1[i].n, vec1[i].s);
    for (long long i = 0; i < k; i++)
      for (long long j = 0; j < k; j++)
        ans += res[i].s * vec2[j].n + res[i].n * vec2[j].s -
               res[i].n * vec2[j].n * ((i != 0) && (j != 0) && (i + j <= k));
    for (long long i = 0; i < k; i++)
      res[i] = P(res[i].n + vec2[i].n, res[i].s + vec2[i].s);
  }
  for (long long i = 0; i < k; i++) ans += res[i].s;
  res[0] = P(res[0].n + 1, res[0].s);
  return res;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (long long i = 0; i < n - 1; i++) {
    static long long u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, -1);
  cout << ans;
  return 0;
}

```