#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
vector<int> G[maxn];
int totson[maxn], subtree[maxn][10], n, k, u, v;
long long ans;
int subtract(int a, int b) { return ((a - b) % k + k) % k; }
void dfs(int cur, int par, int dept) {
  totson[cur] = subtree[cur][dept % k] = 1;
  for (int i = 0; i < G[cur].size(); i++) {
    int son = G[cur][i];
    if (son == par) continue;
    dfs(son, cur, dept + 1);
    for (int a = 0; a < k; a++) {
      for (int b = 0; b < k; b++) {
        int dis = subtract(a + b, 2 * dept);
        int needs = subtract(k, dis);
        ans += (long long)needs * subtree[cur][a] * subtree[son][b];
      }
    }
    totson[cur] += totson[son];
    for (int j = 0; j < k; j++) subtree[cur][j] += subtree[son][j];
  }
  ans += (long long)totson[cur] * (n - totson[cur]);
}
int main() {
  ans = 0;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, -1, 0);
  ans /= k;
  cout << ans << endl;
}

### Prompt

Your task is to create 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 int maxn = 2e5 + 10;
vector<int> G[maxn];
int totson[maxn], subtree[maxn][10], n, k, u, v;
long long ans;
int subtract(int a, int b) { return ((a - b) % k + k) % k; }
void dfs(int cur, int par, int dept) {
  totson[cur] = subtree[cur][dept % k] = 1;
  for (int i = 0; i < G[cur].size(); i++) {
    int son = G[cur][i];
    if (son == par) continue;
    dfs(son, cur, dept + 1);
    for (int a = 0; a < k; a++) {
      for (int b = 0; b < k; b++) {
        int dis = subtract(a + b, 2 * dept);
        int needs = subtract(k, dis);
        ans += (long long)needs * subtree[cur][a] * subtree[son][b];
      }
    }
    totson[cur] += totson[son];
    for (int j = 0; j < k; j++) subtree[cur][j] += subtree[son][j];
  }
  ans += (long long)totson[cur] * (n - totson[cur]);
}
int main() {
  ans = 0;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, -1, 0);
  ans /= k;
  cout << ans << endl;
}

```