#include <bits/stdc++.h>
using namespace std;
const double sn = 1e-6;
const int siz = 200005;
int n, k;
long long ans;
vector<int> adj[siz];
vector<int> child[siz];
bool vis[siz];
struct cost {
  long long n, moves;
};
cost dp[siz][6];
void makerooted(int i) {
  vis[i] = true;
  int sz = adj[i].size();
  for (int j = 0; j < sz; j++) {
    if (!vis[adj[i][j]]) {
      child[i].push_back(adj[i][j]);
      makerooted(adj[i][j]);
    }
  }
}
void dfs(int i) {
  dp[i][0] = {1, 0};
  int sz = child[i].size();
  for (int j = 0; j < sz; j++) {
    int ch = child[i][j];
    dfs(ch);
    for (int l = 0; l < k; l++) {
      dp[i][(l + 1) % k].n += dp[ch][l].n;
      dp[i][(l + 1) % k].moves += dp[ch][l].moves;
      if ((l + 1) % k == 0) dp[i][0].moves += dp[ch][l].n;
    }
  }
}
void solve(int i, int p) {
  if (p != 0) {
    cost arr[6] = {};
    for (int j = 0; j < k; j++) {
      arr[(j + 1) % k] = dp[i][j];
      if ((j + 1) % k == 0) arr[0].moves += dp[i][j].n;
    }
    for (int j = 0; j < k; j++) {
      arr[j].n = dp[p][j].n - arr[j].n;
      arr[j].moves = dp[p][j].moves - arr[j].moves;
    }
    for (int j = 0; j < k; j++) {
      dp[p][j] = arr[j];
    }
    arr[0].n--;
    for (int j = 0; j < k; j++) {
      dp[i][(j + 1) % k].n += arr[j].n;
      dp[i][(j + 1) % k].moves += arr[j].moves;
      if ((j + 1) % k == 0) dp[i][0].moves += arr[j].n;
    }
  }
  for (int j = 0; j < k; j++) {
    ans += dp[i][j].moves + (bool)j * dp[i][j].n;
  }
  int sz = child[i].size();
  for (int j = 0; j < sz; j++) {
    solve(child[i][j], i);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  makerooted(1);
  dfs(1);
  solve(1, 0);
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Generate 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 double sn = 1e-6;
const int siz = 200005;
int n, k;
long long ans;
vector<int> adj[siz];
vector<int> child[siz];
bool vis[siz];
struct cost {
  long long n, moves;
};
cost dp[siz][6];
void makerooted(int i) {
  vis[i] = true;
  int sz = adj[i].size();
  for (int j = 0; j < sz; j++) {
    if (!vis[adj[i][j]]) {
      child[i].push_back(adj[i][j]);
      makerooted(adj[i][j]);
    }
  }
}
void dfs(int i) {
  dp[i][0] = {1, 0};
  int sz = child[i].size();
  for (int j = 0; j < sz; j++) {
    int ch = child[i][j];
    dfs(ch);
    for (int l = 0; l < k; l++) {
      dp[i][(l + 1) % k].n += dp[ch][l].n;
      dp[i][(l + 1) % k].moves += dp[ch][l].moves;
      if ((l + 1) % k == 0) dp[i][0].moves += dp[ch][l].n;
    }
  }
}
void solve(int i, int p) {
  if (p != 0) {
    cost arr[6] = {};
    for (int j = 0; j < k; j++) {
      arr[(j + 1) % k] = dp[i][j];
      if ((j + 1) % k == 0) arr[0].moves += dp[i][j].n;
    }
    for (int j = 0; j < k; j++) {
      arr[j].n = dp[p][j].n - arr[j].n;
      arr[j].moves = dp[p][j].moves - arr[j].moves;
    }
    for (int j = 0; j < k; j++) {
      dp[p][j] = arr[j];
    }
    arr[0].n--;
    for (int j = 0; j < k; j++) {
      dp[i][(j + 1) % k].n += arr[j].n;
      dp[i][(j + 1) % k].moves += arr[j].moves;
      if ((j + 1) % k == 0) dp[i][0].moves += arr[j].n;
    }
  }
  for (int j = 0; j < k; j++) {
    ans += dp[i][j].moves + (bool)j * dp[i][j].n;
  }
  int sz = child[i].size();
  for (int j = 0; j < sz; j++) {
    solve(child[i][j], i);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  makerooted(1);
  dfs(1);
  solve(1, 0);
  printf("%I64d\n", ans);
  return 0;
}

```