#include <bits/stdc++.h>
using namespace std;
struct st {
  long long cnt;
  long long dp[5];
  long long bc;
  st() {
    cnt = bc = 0;
    for (int i = 0; i < 5; ++i) dp[i] = 0;
  }
};
int was[210000];
long long gans = 0;
int n;
vector<int> eds[210000];
int k;
st dfs1(int v) {
  was[v] = 1;
  st now;
  now.cnt = 1;
  now.dp[0] = 1;
  for (int u : eds[v]) {
    if (was[u]) continue;
    st tmp = dfs1(u);
    rotate(tmp.dp, tmp.dp + k - 1, tmp.dp + k);
    tmp.bc += tmp.dp[0];
    gans += tmp.bc * now.cnt;
    gans += tmp.cnt * now.bc;
    for (int i = 0; i < k; ++i)
      for (int j = 0; j < k; ++j)
        gans += tmp.dp[i] * now.dp[j] * ((i + j + k - 1) / k);
    now.cnt += tmp.cnt;
    now.bc += tmp.bc;
    for (int i = 0; i < k; ++i) now.dp[i] += tmp.dp[i];
  }
  return now;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    eds[a].push_back(b);
    eds[b].push_back(a);
  }
  dfs1(0);
  cout << gans << "\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
struct st {
  long long cnt;
  long long dp[5];
  long long bc;
  st() {
    cnt = bc = 0;
    for (int i = 0; i < 5; ++i) dp[i] = 0;
  }
};
int was[210000];
long long gans = 0;
int n;
vector<int> eds[210000];
int k;
st dfs1(int v) {
  was[v] = 1;
  st now;
  now.cnt = 1;
  now.dp[0] = 1;
  for (int u : eds[v]) {
    if (was[u]) continue;
    st tmp = dfs1(u);
    rotate(tmp.dp, tmp.dp + k - 1, tmp.dp + k);
    tmp.bc += tmp.dp[0];
    gans += tmp.bc * now.cnt;
    gans += tmp.cnt * now.bc;
    for (int i = 0; i < k; ++i)
      for (int j = 0; j < k; ++j)
        gans += tmp.dp[i] * now.dp[j] * ((i + j + k - 1) / k);
    now.cnt += tmp.cnt;
    now.bc += tmp.bc;
    for (int i = 0; i < k; ++i) now.dp[i] += tmp.dp[i];
  }
  return now;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    eds[a].push_back(b);
    eds[b].push_back(a);
  }
  dfs1(0);
  cout << gans << "\n";
  return 0;
}

```