#include <bits/stdc++.h>
using namespace std;
int N, K;
vector<int> adj[200005];
int dep[200005];
int par[200005];
int sz[200005];
long long cnttot[5];
long long dp[200005][5];
long long pass[200005][5];
long long add[200005][5];
inline int get(int x) {
  while (x < 0) x += K;
  while (x >= K) x -= K;
  return x;
}
void dfs(int cur) {
  sz[cur] = 1;
  dep[cur] = dep[par[cur]] + 1;
  pass[cur][0] = 1;
  long long c[5];
  for (int i = 0; i < K; i++) c[i] = 0;
  for (int i = 0; i < adj[cur].size(); i++) {
    int nxt = adj[cur][i];
    if (nxt == par[cur]) continue;
    par[nxt] = cur;
    dfs(nxt);
    sz[cur] += sz[nxt];
    for (int j = 0; j <= K; j++) c[j] = c[j] + pass[nxt][get(j - 1)];
  }
  for (int j = 0; j < K; j++) pass[cur][j] += c[j];
  for (int i = 0; i < adj[cur].size(); i++) {
    int nxt = adj[cur][i];
    if (nxt == par[cur]) continue;
    for (int j = 0; j < K; j++) c[j] = c[j] - pass[nxt][get(j - 1)];
    for (int j = 0; j < K; j++) {
      for (int k = 0; k < K; k++) {
        add[cur][get(j + k)] =
            add[cur][get(j + k)] + (long long)(c[j] * pass[nxt][get(k - 1)]);
      }
    }
    for (int j = 0; j < K; j++) c[j] = c[j] + pass[nxt][get(j - 1)];
    for (int j = 0; j < K; j++) {
      dp[cur][j] = dp[cur][j] + dp[nxt][j];
    }
  }
  for (int j = 0; j < K; j++) {
    dp[cur][j] = dp[cur][j] + pass[cur][j];
    dp[cur][j] = dp[cur][j] + add[cur][j] / (long long)2;
  }
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> K;
  memset(cnttot, 0, sizeof(cnttot));
  memset(dp, 0, sizeof(dp));
  memset(pass, 0, sizeof(pass));
  memset(add, 0, sizeof(add));
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  par[1] = 1;
  dep[1] = 0;
  dfs(1);
  long long ans = 0;
  for (int i = 1; i < K; i++) cnttot[i] = dp[1][i];
  for (int i = 1; i <= N; i++) {
    long long res = (long long)sz[i];
    res = (long long)(res * (long long)(sz[1] - sz[i]));
    ans += res;
  }
  if (K == 1) {
    cout << ans << endl;
    return 0;
  }
  for (int i = 1; i < K; i++) {
    ans = ans + (long long)((long long)cnttot[i] * (long long)(K - i));
  }
  ans = ans / (long long)K;
  cout << ans << endl;
}

### Prompt

In cpp, your task is to solve 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;
int N, K;
vector<int> adj[200005];
int dep[200005];
int par[200005];
int sz[200005];
long long cnttot[5];
long long dp[200005][5];
long long pass[200005][5];
long long add[200005][5];
inline int get(int x) {
  while (x < 0) x += K;
  while (x >= K) x -= K;
  return x;
}
void dfs(int cur) {
  sz[cur] = 1;
  dep[cur] = dep[par[cur]] + 1;
  pass[cur][0] = 1;
  long long c[5];
  for (int i = 0; i < K; i++) c[i] = 0;
  for (int i = 0; i < adj[cur].size(); i++) {
    int nxt = adj[cur][i];
    if (nxt == par[cur]) continue;
    par[nxt] = cur;
    dfs(nxt);
    sz[cur] += sz[nxt];
    for (int j = 0; j <= K; j++) c[j] = c[j] + pass[nxt][get(j - 1)];
  }
  for (int j = 0; j < K; j++) pass[cur][j] += c[j];
  for (int i = 0; i < adj[cur].size(); i++) {
    int nxt = adj[cur][i];
    if (nxt == par[cur]) continue;
    for (int j = 0; j < K; j++) c[j] = c[j] - pass[nxt][get(j - 1)];
    for (int j = 0; j < K; j++) {
      for (int k = 0; k < K; k++) {
        add[cur][get(j + k)] =
            add[cur][get(j + k)] + (long long)(c[j] * pass[nxt][get(k - 1)]);
      }
    }
    for (int j = 0; j < K; j++) c[j] = c[j] + pass[nxt][get(j - 1)];
    for (int j = 0; j < K; j++) {
      dp[cur][j] = dp[cur][j] + dp[nxt][j];
    }
  }
  for (int j = 0; j < K; j++) {
    dp[cur][j] = dp[cur][j] + pass[cur][j];
    dp[cur][j] = dp[cur][j] + add[cur][j] / (long long)2;
  }
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> K;
  memset(cnttot, 0, sizeof(cnttot));
  memset(dp, 0, sizeof(dp));
  memset(pass, 0, sizeof(pass));
  memset(add, 0, sizeof(add));
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  par[1] = 1;
  dep[1] = 0;
  dfs(1);
  long long ans = 0;
  for (int i = 1; i < K; i++) cnttot[i] = dp[1][i];
  for (int i = 1; i <= N; i++) {
    long long res = (long long)sz[i];
    res = (long long)(res * (long long)(sz[1] - sz[i]));
    ans += res;
  }
  if (K == 1) {
    cout << ans << endl;
    return 0;
  }
  for (int i = 1; i < K; i++) {
    ans = ans + (long long)((long long)cnttot[i] * (long long)(K - i));
  }
  ans = ans / (long long)K;
  cout << ans << endl;
}

```