#include <bits/stdc++.h>
using namespace std;
struct data {
  long long int rem[6], ans, sub, newVal;
};
long long int n, k;
vector<long long int> g[200000 + 5];
data dp[200000 + 5];
data pre() {
  data r;
  int i;
  for (i = 0; i < 5; i++) r.rem[i] = 0;
  r.sub = 1;
  r.ans = 0;
  r.newVal = 0;
  return r;
}
data dfs(long long int u, long long int p) {
  long long int i, v, j, val, l, f = 0;
  data r, x;
  long long int rem[6];
  r = pre();
  x = pre();
  x.sub = 0;
  for (i = 0; i < (long long int)g[u].size(); i++) {
    v = g[u][i];
    if (v == p) continue;
    dp[v] = dfs(v, u);
    r.ans += dp[v].ans;
    r.sub += dp[v].sub;
    val = dp[v].newVal;
    val += dp[v].rem[0] + 1;
    for (j = 0; j < k; j++) rem[j] = dp[v].rem[j];
    for (j = 0; j < k; j++) dp[v].rem[(j + 1) % k] = rem[j];
    dp[v].rem[1 % k]++;
    for (j = 0; j < k; j++) x.rem[j] += dp[v].rem[j];
    dp[v].ans = val;
    r.ans += val;
    r.newVal += val;
    x.sub += dp[v].sub;
    x.ans += val;
  }
  for (i = 0; i < (long long int)g[u].size(); i++) {
    v = g[u][i];
    if (v == p) continue;
    x.ans -= dp[v].ans;
    x.sub -= dp[v].sub;
    for (j = 0; j < k; j++) x.rem[j] -= dp[v].rem[j];
    r.ans += dp[v].ans * x.sub;
    for (j = 1; j < k; j++)
      for (l = 1; l < k; l++)
        if ((j + l) <= k) f += dp[v].rem[j] * x.rem[l];
    x.ans += dp[v].ans;
    x.sub += dp[v].sub;
    for (j = 0; j < k; j++) {
      x.rem[j] += dp[v].rem[j];
      r.rem[j] += dp[v].rem[j];
    }
  }
  f /= 2;
  r.ans -= f;
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int i, u, v;
  cin >> n >> k;
  for (i = 1; i <= n; i++) dp[i] = pre();
  for (i = 1; i < n; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  cout << dfs(1, 1).ans;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 data {
  long long int rem[6], ans, sub, newVal;
};
long long int n, k;
vector<long long int> g[200000 + 5];
data dp[200000 + 5];
data pre() {
  data r;
  int i;
  for (i = 0; i < 5; i++) r.rem[i] = 0;
  r.sub = 1;
  r.ans = 0;
  r.newVal = 0;
  return r;
}
data dfs(long long int u, long long int p) {
  long long int i, v, j, val, l, f = 0;
  data r, x;
  long long int rem[6];
  r = pre();
  x = pre();
  x.sub = 0;
  for (i = 0; i < (long long int)g[u].size(); i++) {
    v = g[u][i];
    if (v == p) continue;
    dp[v] = dfs(v, u);
    r.ans += dp[v].ans;
    r.sub += dp[v].sub;
    val = dp[v].newVal;
    val += dp[v].rem[0] + 1;
    for (j = 0; j < k; j++) rem[j] = dp[v].rem[j];
    for (j = 0; j < k; j++) dp[v].rem[(j + 1) % k] = rem[j];
    dp[v].rem[1 % k]++;
    for (j = 0; j < k; j++) x.rem[j] += dp[v].rem[j];
    dp[v].ans = val;
    r.ans += val;
    r.newVal += val;
    x.sub += dp[v].sub;
    x.ans += val;
  }
  for (i = 0; i < (long long int)g[u].size(); i++) {
    v = g[u][i];
    if (v == p) continue;
    x.ans -= dp[v].ans;
    x.sub -= dp[v].sub;
    for (j = 0; j < k; j++) x.rem[j] -= dp[v].rem[j];
    r.ans += dp[v].ans * x.sub;
    for (j = 1; j < k; j++)
      for (l = 1; l < k; l++)
        if ((j + l) <= k) f += dp[v].rem[j] * x.rem[l];
    x.ans += dp[v].ans;
    x.sub += dp[v].sub;
    for (j = 0; j < k; j++) {
      x.rem[j] += dp[v].rem[j];
      r.rem[j] += dp[v].rem[j];
    }
  }
  f /= 2;
  r.ans -= f;
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int i, u, v;
  cin >> n >> k;
  for (i = 1; i <= n; i++) dp[i] = pre();
  for (i = 1; i < n; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  cout << dfs(1, 1).ans;
  return 0;
}

```