#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
const int N = 200010;
int las[N], nxt[N << 1], to[N << 1], cnt;
inline void add(int u, int v) {
  cnt++;
  to[cnt] = v;
  nxt[cnt] = las[u];
  las[u] = cnt;
}
inline void add2() {
  static int a, b;
  scanf("%d%d", &a, &b);
  add(a, b);
  add(b, a);
}
int n, k;
long long a[N][7], ans;
void dfs1(int now, int la) {
  int i, j;
  a[now][1 % k] = 1;
  a[now][k] = 1;
  for (i = las[now]; i; i = nxt[i])
    if (to[i] != la) {
      dfs1(to[i], now);
      for (j = 0; j < k; j++) a[now][(j + 1) % k] += a[to[i]][j];
      a[now][k] += a[to[i]][k] + a[to[i]][0];
    }
}
inline void swap(long long &x, long long &y) { x ^= y ^= x ^= y; }
void dfs2(int now, int la) {
  int i, j;
  long long tmp;
  if (la) {
    for (i = 0; i <= k; i++) a[now][i] = a[la][i] - a[now][i];
    a[now][k] += a[now][0];
    tmp = a[now][k - 1];
    for (i = 0; i < k; i++) swap(a[now][i], tmp);
  } else
    for (i = 0; i <= k; i++) a[now][i] = 0;
  a[now][0]++;
  for (i = las[now]; i; i = nxt[i])
    if (to[i] != la)
      for (j = 0; j <= k; j++) a[now][j] += a[to[i]][j];
  ans += a[now][k];
  for (i = las[now]; i; i = nxt[i])
    if (to[i] != la) dfs2(to[i], now);
}
int main() {
  int i;
  scanf("%d%d", &n, &k);
  for (i = 1; i < n; i++) add2();
  dfs1(1, 0);
  dfs2(1, 0);
  printf("%lld\n", ans >> 1);
  return 0;
}

### 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>
#pragma GCC optimize("Ofast")
const int N = 200010;
int las[N], nxt[N << 1], to[N << 1], cnt;
inline void add(int u, int v) {
  cnt++;
  to[cnt] = v;
  nxt[cnt] = las[u];
  las[u] = cnt;
}
inline void add2() {
  static int a, b;
  scanf("%d%d", &a, &b);
  add(a, b);
  add(b, a);
}
int n, k;
long long a[N][7], ans;
void dfs1(int now, int la) {
  int i, j;
  a[now][1 % k] = 1;
  a[now][k] = 1;
  for (i = las[now]; i; i = nxt[i])
    if (to[i] != la) {
      dfs1(to[i], now);
      for (j = 0; j < k; j++) a[now][(j + 1) % k] += a[to[i]][j];
      a[now][k] += a[to[i]][k] + a[to[i]][0];
    }
}
inline void swap(long long &x, long long &y) { x ^= y ^= x ^= y; }
void dfs2(int now, int la) {
  int i, j;
  long long tmp;
  if (la) {
    for (i = 0; i <= k; i++) a[now][i] = a[la][i] - a[now][i];
    a[now][k] += a[now][0];
    tmp = a[now][k - 1];
    for (i = 0; i < k; i++) swap(a[now][i], tmp);
  } else
    for (i = 0; i <= k; i++) a[now][i] = 0;
  a[now][0]++;
  for (i = las[now]; i; i = nxt[i])
    if (to[i] != la)
      for (j = 0; j <= k; j++) a[now][j] += a[to[i]][j];
  ans += a[now][k];
  for (i = las[now]; i; i = nxt[i])
    if (to[i] != la) dfs2(to[i], now);
}
int main() {
  int i;
  scanf("%d%d", &n, &k);
  for (i = 1; i < n; i++) add2();
  dfs1(1, 0);
  dfs2(1, 0);
  printf("%lld\n", ans >> 1);
  return 0;
}

```