#include <bits/stdc++.h>
using namespace std;
long long head[200100], nxt[200100 << 1], ver[200100 << 1], tot;
inline void addedge(long long a, long long b) {
  nxt[++tot] = head[a];
  ver[tot] = b;
  head[a] = tot;
  nxt[++tot] = head[b];
  ver[tot] = a;
  head[b] = tot;
}
long long f[200100][5], s[200100];
long long ans = 0;
long long n, k;
inline void dfs(long long x, long long fat) {
  f[x][0] = 1;
  for (long long i = head[x]; i; i = nxt[i]) {
    long long y = ver[i];
    if (y == fat) continue;
    dfs(y, x);
    s[x] += s[y] + f[y][0];
    for (long long j = 0; j < k; j++) f[x][(j + 1) % k] += f[y][j];
  }
  ans += s[x];
}
long long F[200100][5], S[200100];
inline void getans(long long x, long long fat) {
  if (x != 1) {
    for (long long j = 0; j < k; j++) F[x][(j + 1) % k] += F[fat][j];
    S[x] = S[fat] + F[fat][0];
    ans += S[x];
  }
  S[x] += s[x];
  for (long long j = 0; j < k; j++) F[x][j] += f[x][j];
  for (long long i = head[x]; i; i = nxt[i]) {
    long long y = ver[i];
    if (y == fat) continue;
    S[x] -= s[y] + f[y][0];
    for (long long j = 0; j < k; j++) F[x][(j + 1) % k] -= f[y][j];
    getans(y, x);
    for (long long j = 0; j < k; j++) F[x][(j + 1) % k] += f[y][j];
    S[x] += s[y] + f[y][0];
  }
}
inline void init() {
  scanf("%lld%lld", &n, &k);
  for (long long i = 1, a, b; i < n; i++)
    scanf("%lld%lld", &a, &b), addedge(a, b);
}
signed main() {
  init();
  dfs(1, 0);
  getans(1, 0);
  printf("%lld", ans >> 1);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long head[200100], nxt[200100 << 1], ver[200100 << 1], tot;
inline void addedge(long long a, long long b) {
  nxt[++tot] = head[a];
  ver[tot] = b;
  head[a] = tot;
  nxt[++tot] = head[b];
  ver[tot] = a;
  head[b] = tot;
}
long long f[200100][5], s[200100];
long long ans = 0;
long long n, k;
inline void dfs(long long x, long long fat) {
  f[x][0] = 1;
  for (long long i = head[x]; i; i = nxt[i]) {
    long long y = ver[i];
    if (y == fat) continue;
    dfs(y, x);
    s[x] += s[y] + f[y][0];
    for (long long j = 0; j < k; j++) f[x][(j + 1) % k] += f[y][j];
  }
  ans += s[x];
}
long long F[200100][5], S[200100];
inline void getans(long long x, long long fat) {
  if (x != 1) {
    for (long long j = 0; j < k; j++) F[x][(j + 1) % k] += F[fat][j];
    S[x] = S[fat] + F[fat][0];
    ans += S[x];
  }
  S[x] += s[x];
  for (long long j = 0; j < k; j++) F[x][j] += f[x][j];
  for (long long i = head[x]; i; i = nxt[i]) {
    long long y = ver[i];
    if (y == fat) continue;
    S[x] -= s[y] + f[y][0];
    for (long long j = 0; j < k; j++) F[x][(j + 1) % k] -= f[y][j];
    getans(y, x);
    for (long long j = 0; j < k; j++) F[x][(j + 1) % k] += f[y][j];
    S[x] += s[y] + f[y][0];
  }
}
inline void init() {
  scanf("%lld%lld", &n, &k);
  for (long long i = 1, a, b; i < n; i++)
    scanf("%lld%lld", &a, &b), addedge(a, b);
}
signed main() {
  init();
  dfs(1, 0);
  getans(1, 0);
  printf("%lld", ans >> 1);
  return 0;
}

```