#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T sqr(T a) {
  return a * a;
};
const double pi = acos(-1.0);
const int INF = INT_MAX;
const long long LLINF = LLONG_MAX;
const int MAX_N = 2e5 + 10;
vector<int> G[MAX_N];
int count_subtree[MAX_N][5];
int total_subtree[MAX_N];
long long ans;
int n, k;
int subtract(int a, int b) { return ((a - b) % k + k) % k; }
void dfs(int v, int p, int d) {
  count_subtree[v][d % k] = total_subtree[v] = 1;
  for (int u : G[v])
    if (u != p) {
      dfs(u, v, d + 1);
      for (int i = 0; i < k; i++) {
        for (int j = 0; j < k; j++) {
          int reminder = subtract(i + j, 2 * d);
          int needs = subtract(k, reminder);
          ans += (long long)needs * count_subtree[v][i] * count_subtree[u][j];
        }
      }
      for (int i = 0; i < k; i++) count_subtree[v][i] += count_subtree[u][i];
      total_subtree[v] += total_subtree[u];
    }
  ans += (long long)total_subtree[v] * (n - total_subtree[v]);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(1, -1, 0);
  cout << (ans / k) << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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;
template <typename T>
inline T sqr(T a) {
  return a * a;
};
const double pi = acos(-1.0);
const int INF = INT_MAX;
const long long LLINF = LLONG_MAX;
const int MAX_N = 2e5 + 10;
vector<int> G[MAX_N];
int count_subtree[MAX_N][5];
int total_subtree[MAX_N];
long long ans;
int n, k;
int subtract(int a, int b) { return ((a - b) % k + k) % k; }
void dfs(int v, int p, int d) {
  count_subtree[v][d % k] = total_subtree[v] = 1;
  for (int u : G[v])
    if (u != p) {
      dfs(u, v, d + 1);
      for (int i = 0; i < k; i++) {
        for (int j = 0; j < k; j++) {
          int reminder = subtract(i + j, 2 * d);
          int needs = subtract(k, reminder);
          ans += (long long)needs * count_subtree[v][i] * count_subtree[u][j];
        }
      }
      for (int i = 0; i < k; i++) count_subtree[v][i] += count_subtree[u][i];
      total_subtree[v] += total_subtree[u];
    }
  ans += (long long)total_subtree[v] * (n - total_subtree[v]);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(1, -1, 0);
  cout << (ans / k) << endl;
  return 0;
}

```