#include <bits/stdc++.h>
using namespace std;
using unlong = unsigned long long;
int n, k;
vector<int> adj[200007];
int siz[200007];
int cnt[200007][5];
long long answer;
int sub(int x, int y) { return ((x - y) % k + k) % k; }
void dfs(int u = 1, int p = 0, int depth = 0) {
  siz[u] = 1;
  cnt[u][depth % k] = 1;
  for (int v : adj[u]) {
    if (v == p) continue;
    dfs(v, u, depth + 1);
    for (int i = 0; i <= k - 1; i++)
      for (int j = 0; j <= k - 1; j++) {
        int rem = sub(i + j, depth << 1);
        int ned = sub(k, rem);
        answer += (long long)ned * cnt[u][i] * cnt[v][j];
      }
    for (int i = 0; i <= k - 1; i++) cnt[u][i] += cnt[v][i];
    siz[u] += siz[v];
  }
  answer += (long long)siz[u] * (n - siz[u]);
}
void enter() {
  cin >> n >> k;
  int u, v;
  for (int i = 2; i <= n; i++)
    cin >> u >> v, adj[u].push_back(v), adj[v].push_back(u);
}
void solve() {
  dfs();
  assert(answer % k == 0);
  cout << answer / k << "\n";
}
int main() {
  {
    if (fopen("spoj"
              ".inp",
              "r"))
      freopen(
          "spoj"
          ".inp",
          "r", stdin),
          freopen(
              "spoj"
              ".out",
              "w", stdout);
  };
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  enter();
  solve();
  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;
using unlong = unsigned long long;
int n, k;
vector<int> adj[200007];
int siz[200007];
int cnt[200007][5];
long long answer;
int sub(int x, int y) { return ((x - y) % k + k) % k; }
void dfs(int u = 1, int p = 0, int depth = 0) {
  siz[u] = 1;
  cnt[u][depth % k] = 1;
  for (int v : adj[u]) {
    if (v == p) continue;
    dfs(v, u, depth + 1);
    for (int i = 0; i <= k - 1; i++)
      for (int j = 0; j <= k - 1; j++) {
        int rem = sub(i + j, depth << 1);
        int ned = sub(k, rem);
        answer += (long long)ned * cnt[u][i] * cnt[v][j];
      }
    for (int i = 0; i <= k - 1; i++) cnt[u][i] += cnt[v][i];
    siz[u] += siz[v];
  }
  answer += (long long)siz[u] * (n - siz[u]);
}
void enter() {
  cin >> n >> k;
  int u, v;
  for (int i = 2; i <= n; i++)
    cin >> u >> v, adj[u].push_back(v), adj[v].push_back(u);
}
void solve() {
  dfs();
  assert(answer % k == 0);
  cout << answer / k << "\n";
}
int main() {
  {
    if (fopen("spoj"
              ".inp",
              "r"))
      freopen(
          "spoj"
          ".inp",
          "r", stdin),
          freopen(
              "spoj"
              ".out",
              "w", stdout);
  };
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  enter();
  solve();
  return 0;
}

```