#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 200001, MAX_K = 5;
int N, K;
int num[MAX_N][MAX_K];
long long sum[MAX_N][MAX_K];
vector<int> adj[MAX_N];
vector<int> info_num[MAX_N][MAX_K];
vector<long long> info_sum[MAX_N][MAX_K];
void dfs1(int v, int dad) {
  num[v][0] = 1;
  for (int i = 0; i < (int)adj[v].size(); i++) {
    int u = adj[v][i];
    if (u == dad) continue;
    dfs1(u, v);
    for (int k = 0; k < K; k++) {
      info_num[v][k][i] = num[u][(k + K - 1) % K];
      info_sum[v][k][i] = sum[u][(k + K - 1) % K];
    }
    info_sum[v][(K == 1) ? 0 : 1][i] += (long long)num[u][0];
    for (int k = 0; k < K; k++) {
      num[v][k] += info_num[v][k][i];
      sum[v][k] += info_sum[v][k][i];
    }
  }
}
void dfs2(int v, int idx, int dad) {
  if (dad != v) {
    for (int i = 0; i < K; i++) {
      num[v][i] +=
          num[dad][(i + K - 1) % K] - info_num[dad][(i + K - 1) % K][idx];
      sum[v][i] +=
          sum[dad][(i + K - 1) % K] - info_sum[dad][(i + K - 1) % K][idx];
    }
    sum[v][(K == 1) ? 0 : 1] +=
        (long long)(num[dad][0] - info_num[dad][0][idx]);
  }
  for (int i = 0; i < (int)adj[v].size(); i++) {
    int u = adj[v][i];
    if (u == dad) continue;
    dfs2(u, i, v);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> K;
  for (int i = 0; i < N - 1; i++) {
    int A, B;
    cin >> A >> B;
    adj[A].push_back(B);
    adj[B].push_back(A);
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 0; j < K; j++) {
      info_num[i][j].resize(adj[i].size());
      info_sum[i][j].resize(adj[i].size());
    }
  }
  dfs1(1, 1);
  dfs2(1, 0, 1);
  long long ans = 0;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j < K; j++) ans += sum[i][j];
  cout << ans / 2LL << "\n";
  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;
const int MAX_N = 200001, MAX_K = 5;
int N, K;
int num[MAX_N][MAX_K];
long long sum[MAX_N][MAX_K];
vector<int> adj[MAX_N];
vector<int> info_num[MAX_N][MAX_K];
vector<long long> info_sum[MAX_N][MAX_K];
void dfs1(int v, int dad) {
  num[v][0] = 1;
  for (int i = 0; i < (int)adj[v].size(); i++) {
    int u = adj[v][i];
    if (u == dad) continue;
    dfs1(u, v);
    for (int k = 0; k < K; k++) {
      info_num[v][k][i] = num[u][(k + K - 1) % K];
      info_sum[v][k][i] = sum[u][(k + K - 1) % K];
    }
    info_sum[v][(K == 1) ? 0 : 1][i] += (long long)num[u][0];
    for (int k = 0; k < K; k++) {
      num[v][k] += info_num[v][k][i];
      sum[v][k] += info_sum[v][k][i];
    }
  }
}
void dfs2(int v, int idx, int dad) {
  if (dad != v) {
    for (int i = 0; i < K; i++) {
      num[v][i] +=
          num[dad][(i + K - 1) % K] - info_num[dad][(i + K - 1) % K][idx];
      sum[v][i] +=
          sum[dad][(i + K - 1) % K] - info_sum[dad][(i + K - 1) % K][idx];
    }
    sum[v][(K == 1) ? 0 : 1] +=
        (long long)(num[dad][0] - info_num[dad][0][idx]);
  }
  for (int i = 0; i < (int)adj[v].size(); i++) {
    int u = adj[v][i];
    if (u == dad) continue;
    dfs2(u, i, v);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> K;
  for (int i = 0; i < N - 1; i++) {
    int A, B;
    cin >> A >> B;
    adj[A].push_back(B);
    adj[B].push_back(A);
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 0; j < K; j++) {
      info_num[i][j].resize(adj[i].size());
      info_sum[i][j].resize(adj[i].size());
    }
  }
  dfs1(1, 1);
  dfs2(1, 0, 1);
  long long ans = 0;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j < K; j++) ans += sum[i][j];
  cout << ans / 2LL << "\n";
  return 0;
}

```