#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
const int maxm = 2 * maxn;
int n, m, u, v, k;
struct edge {
  int v, next;
} e[maxm];
int h[maxn], num;
long long f[maxn][6];
long long sum[maxn][6];
long long ans = 0;
void build(int u, int v) {
  num++;
  e[num].v = v;
  e[num].next = h[u];
  h[u] = num;
}
void add(int x, int v) {
  long long temp[6], s[6];
  for (int i = 0; i <= k; i++) temp[i] = f[x][i];
  for (int i = 0; i < k; i++) ans += f[x][i] * f[v][k];
  for (int i = 0; i < k; i++)
    for (int j = 1; j < k; j++) ans -= f[x][i] * f[v][j];
  for (int i = 0; i < k; i++) {
    memset(s, 0, sizeof(s));
    s[0] = temp[k - 1];
    ans += f[v][i] * temp[0];
    ans += f[v][i] * temp[k];
    for (int j = 1; j < k; j++) s[j] += temp[j - 1];
    temp[k] += temp[0];
    for (int j = 0; j < k; j++) temp[j] = s[j];
  }
}
void dfs(int x, int fa) {
  f[x][0]++;
  for (int i = h[x]; i; i = e[i].next) {
    if (e[i].v == fa) continue;
    dfs(e[i].v, x);
    add(x, e[i].v);
    f[x][0] += f[e[i].v][k - 1];
    f[x][k] += f[e[i].v][0];
    f[x][k] += f[e[i].v][k];
    for (int j = 1; j < k; j++) f[x][j] += f[e[i].v][j - 1];
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    build(u, v);
    build(v, u);
  }
  dfs(1, 0);
  cout << ans << endl;
  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;
const int maxn = 200010;
const int maxm = 2 * maxn;
int n, m, u, v, k;
struct edge {
  int v, next;
} e[maxm];
int h[maxn], num;
long long f[maxn][6];
long long sum[maxn][6];
long long ans = 0;
void build(int u, int v) {
  num++;
  e[num].v = v;
  e[num].next = h[u];
  h[u] = num;
}
void add(int x, int v) {
  long long temp[6], s[6];
  for (int i = 0; i <= k; i++) temp[i] = f[x][i];
  for (int i = 0; i < k; i++) ans += f[x][i] * f[v][k];
  for (int i = 0; i < k; i++)
    for (int j = 1; j < k; j++) ans -= f[x][i] * f[v][j];
  for (int i = 0; i < k; i++) {
    memset(s, 0, sizeof(s));
    s[0] = temp[k - 1];
    ans += f[v][i] * temp[0];
    ans += f[v][i] * temp[k];
    for (int j = 1; j < k; j++) s[j] += temp[j - 1];
    temp[k] += temp[0];
    for (int j = 0; j < k; j++) temp[j] = s[j];
  }
}
void dfs(int x, int fa) {
  f[x][0]++;
  for (int i = h[x]; i; i = e[i].next) {
    if (e[i].v == fa) continue;
    dfs(e[i].v, x);
    add(x, e[i].v);
    f[x][0] += f[e[i].v][k - 1];
    f[x][k] += f[e[i].v][0];
    f[x][k] += f[e[i].v][k];
    for (int j = 1; j < k; j++) f[x][j] += f[e[i].v][j - 1];
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    build(u, v);
    build(v, u);
  }
  dfs(1, 0);
  cout << ans << endl;
  return 0;
}

```