#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 5;
const int MAXK = 10;
vector<int> G[MAXN];
int n, a, b, zm, k, tab[MAXN][MAXK], tab2[MAXN][MAXK];
long long wyn1[MAXN], wyn2[MAXN], pom[MAXK];
long long wynik;
void dfs1(int, int);
void dfs2(int, int);
int main(void) {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs1(1, 0);
  for (int i = 0; i < k; i++) tab2[1][i] = tab[1][i];
  wyn2[1] = wyn1[1];
  dfs2(1, 0);
  for (int i = 1; i <= n; i++) {
    wynik += wyn2[i];
  }
  cout << wynik / 2;
}
void dfs1(int v, int oj) {
  for (int i = 0; i < G[v].size(); i++) {
    int w = G[v][i];
    if (w == oj) continue;
    dfs1(w, v);
    for (int i = 0; i < k; i++) {
      tab[v][(i + 1) % k] += tab[w][i];
    }
    wyn1[v] += tab[w][k - 1];
    wyn1[v] += wyn1[w];
    tab[v][0]++;
    wyn1[v]++;
  }
}
void dfs2(int v, int oj) {
  for (int i = 0; i < G[v].size(); i++) {
    int w = G[v][i];
    if (w == oj) continue;
    for (int i = 0; i < 10; i++) pom[i] = 0;
    for (int i = 0; i < k; i++) {
      pom[i] = tab2[v][i];
      pom[i] -= tab[w][(i + k - 1) % k];
    }
    pom[0]--;
    for (int i = 0; i < k; i++) {
      tab2[w][(i + 1) % k] += pom[i];
      tab2[w][i] += tab[w][i];
    }
    tab2[w][0]++;
    wyn2[w] = wyn2[v];
    wyn2[w] += pom[k - 1];
    wyn2[w] -= tab[w][k - 1];
    dfs2(w, v);
  }
}

### Prompt

Please provide a cpp coded solution 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;
const int MAXN = 1e6 + 5;
const int MAXK = 10;
vector<int> G[MAXN];
int n, a, b, zm, k, tab[MAXN][MAXK], tab2[MAXN][MAXK];
long long wyn1[MAXN], wyn2[MAXN], pom[MAXK];
long long wynik;
void dfs1(int, int);
void dfs2(int, int);
int main(void) {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs1(1, 0);
  for (int i = 0; i < k; i++) tab2[1][i] = tab[1][i];
  wyn2[1] = wyn1[1];
  dfs2(1, 0);
  for (int i = 1; i <= n; i++) {
    wynik += wyn2[i];
  }
  cout << wynik / 2;
}
void dfs1(int v, int oj) {
  for (int i = 0; i < G[v].size(); i++) {
    int w = G[v][i];
    if (w == oj) continue;
    dfs1(w, v);
    for (int i = 0; i < k; i++) {
      tab[v][(i + 1) % k] += tab[w][i];
    }
    wyn1[v] += tab[w][k - 1];
    wyn1[v] += wyn1[w];
    tab[v][0]++;
    wyn1[v]++;
  }
}
void dfs2(int v, int oj) {
  for (int i = 0; i < G[v].size(); i++) {
    int w = G[v][i];
    if (w == oj) continue;
    for (int i = 0; i < 10; i++) pom[i] = 0;
    for (int i = 0; i < k; i++) {
      pom[i] = tab2[v][i];
      pom[i] -= tab[w][(i + k - 1) % k];
    }
    pom[0]--;
    for (int i = 0; i < k; i++) {
      tab2[w][(i + 1) % k] += pom[i];
      tab2[w][i] += tab[w][i];
    }
    tab2[w][0]++;
    wyn2[w] = wyn2[v];
    wyn2[w] += pom[k - 1];
    wyn2[w] -= tab[w][k - 1];
    dfs2(w, v);
  }
}

```