#include <bits/stdc++.h>
const long long mod = 1e9 + 7;
using namespace std;
const long long maxn = 2e5 + 10;
long long n, k;
struct edge {
  long long to, nxt;
} e[maxn * 2];
long long head[maxn], cnt;
long long bac[maxn][6], size[maxn], num[maxn];
long long ans;
void add(long long x, long long y) {
  e[++cnt] = (edge){y, head[x]};
  head[x] = cnt;
}
void dfs(long long x, long long f) {
  if (x == 9) x = 9;
  size[x] = 1;
  for (long long i = head[x]; i; i = e[i].nxt) {
    long long u = e[i].to;
    if (u == f) continue;
    dfs(u, x);
    if (u == 12) u = 12;
    ans += num[x] * size[u] + (size[x] - 1) * num[u];
    size[x] += size[u];
    num[x] += num[u];
    for (long long i = 0; i < k; i++) {
      for (long long j = 0; j < k; j++) {
        long long cnn = bac[x][i] * bac[u][j];
        if ((i + j) % k != 0) ans += cnn;
        ans += cnn * ((i + j) / k);
      }
      bac[x][i] += bac[u][i];
    }
  }
  ans += num[x];
  for (long long i = 1; i < k; i++) ans += bac[x][i];
  if (k == 5) {
    long long tmp = bac[x][0];
    bac[x][0] = bac[x][4];
    num[x] += bac[x][4];
    bac[x][4] = bac[x][3];
    bac[x][3] = bac[x][2];
    bac[x][2] = bac[x][1];
    bac[x][1] = tmp + 1;
  } else if (k == 4) {
    long long tmp = bac[x][0];
    bac[x][0] = bac[x][3];
    num[x] += bac[x][3];
    bac[x][3] = bac[x][2];
    bac[x][2] = bac[x][1];
    bac[x][1] = tmp + 1;
  } else if (k == 3) {
    long long tmp = bac[x][0];
    bac[x][0] = bac[x][2];
    num[x] += bac[x][2];
    bac[x][2] = bac[x][1];
    bac[x][1] = tmp + 1;
  } else if (k == 2) {
    long long tmp = bac[x][0];
    bac[x][0] = bac[x][1];
    num[x] += bac[x][1];
    bac[x][1] = tmp + 1;
  } else {
    bac[x][0] += size[x];
    num[x] += size[x];
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    add(x, y);
    add(y, x);
  }
  dfs(1, 0);
  cout << ans << endl;
  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>
const long long mod = 1e9 + 7;
using namespace std;
const long long maxn = 2e5 + 10;
long long n, k;
struct edge {
  long long to, nxt;
} e[maxn * 2];
long long head[maxn], cnt;
long long bac[maxn][6], size[maxn], num[maxn];
long long ans;
void add(long long x, long long y) {
  e[++cnt] = (edge){y, head[x]};
  head[x] = cnt;
}
void dfs(long long x, long long f) {
  if (x == 9) x = 9;
  size[x] = 1;
  for (long long i = head[x]; i; i = e[i].nxt) {
    long long u = e[i].to;
    if (u == f) continue;
    dfs(u, x);
    if (u == 12) u = 12;
    ans += num[x] * size[u] + (size[x] - 1) * num[u];
    size[x] += size[u];
    num[x] += num[u];
    for (long long i = 0; i < k; i++) {
      for (long long j = 0; j < k; j++) {
        long long cnn = bac[x][i] * bac[u][j];
        if ((i + j) % k != 0) ans += cnn;
        ans += cnn * ((i + j) / k);
      }
      bac[x][i] += bac[u][i];
    }
  }
  ans += num[x];
  for (long long i = 1; i < k; i++) ans += bac[x][i];
  if (k == 5) {
    long long tmp = bac[x][0];
    bac[x][0] = bac[x][4];
    num[x] += bac[x][4];
    bac[x][4] = bac[x][3];
    bac[x][3] = bac[x][2];
    bac[x][2] = bac[x][1];
    bac[x][1] = tmp + 1;
  } else if (k == 4) {
    long long tmp = bac[x][0];
    bac[x][0] = bac[x][3];
    num[x] += bac[x][3];
    bac[x][3] = bac[x][2];
    bac[x][2] = bac[x][1];
    bac[x][1] = tmp + 1;
  } else if (k == 3) {
    long long tmp = bac[x][0];
    bac[x][0] = bac[x][2];
    num[x] += bac[x][2];
    bac[x][2] = bac[x][1];
    bac[x][1] = tmp + 1;
  } else if (k == 2) {
    long long tmp = bac[x][0];
    bac[x][0] = bac[x][1];
    num[x] += bac[x][1];
    bac[x][1] = tmp + 1;
  } else {
    bac[x][0] += size[x];
    num[x] += size[x];
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    add(x, y);
    add(y, x);
  }
  dfs(1, 0);
  cout << ans << endl;
  return 0;
}

```