#include <bits/stdc++.h>
using namespace std;
using namespace std;
const int K = 7;
const int N = 200500;
int n, k;
int sz[K][N], ups[K][N];
long long len[K][N], ways[K][N];
long long f[K][N], fuh[K][N], f1[K][N], fup[K][N];
vector<int> g[N];
void dfs(int ver, int prev) {
  f[0][ver] = 1;
  sz[0][ver] = 1;
  for (int i = 0; i < (int)g[ver].size(); i++) {
    int to = g[ver][i];
    if (to != prev) {
      dfs(to, ver);
      for (int j = 0; j < k; j++) {
        sz[(j + 1) % k][ver] += sz[j][to];
        f[(j + 1) % k][ver] += f[j][to];
        f1[(j + 1) % k][ver] += (f1[j][to] + 1LL * sz[j][to]);
      }
    }
  }
}
void dfs_up(int ver, int prev) {
  ++fup[0][ver];
  ++ups[0][ver];
  long long all[k];
  long long kek[k];
  int ps[k];
  for (int j = 0; j < k; ++j) ps[j] = 0;
  for (int j = 0; j < k; ++j) all[j] = kek[j] = 0LL;
  for (int i = 0; i < (int)g[ver].size(); i++) {
    int to = g[ver][i];
    if (to != prev) {
      for (int j = 0; j < k; j++) {
        all[j] += f[j][to];
        ps[j] += sz[j][to];
        kek[j] += f1[j][to];
      }
    }
  }
  for (int i = 0; i < (int)g[ver].size(); i++) {
    int to = g[ver][i];
    if (to == prev) {
      continue;
    }
    for (int j = 0; j < k; j++) {
      all[j] -= f[j][to];
      ps[j] -= sz[j][to];
      kek[j] -= f1[j][to];
    }
    for (int j = 0; j < k; j++) {
      fup[(j + 2) % k][to] += all[j];
      fup[(j + 1) % k][to] += fup[j][ver];
      fuh[(j + 2) % k][to] += (kek[j] + 2LL * ps[j]);
      fuh[(j + 1) % k][to] += (fuh[j][ver] + 1LL * ups[j][ver]);
      ups[(j + 1) % k][to] += ups[j][ver];
      ups[(j + 2) % k][to] += ps[j];
    }
    for (int j = 0; j < k; j++) {
      all[j] += f[j][to];
      ps[j] += sz[j][to];
      kek[j] += f1[j][to];
    }
  }
  for (int i = 0; i < (int)g[ver].size(); i++) {
    int to = g[ver][i];
    if (to != prev) {
      dfs_up(to, ver);
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 1);
  dfs_up(1, 1);
  long long ans = 0LL;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < k; j++) {
      ways[j][i] = f[j][i] + fup[j][i];
      len[j][i] = f1[j][i] + fuh[j][i];
      ans =
          ans + (len[j][i] - ways[j][i] * j) / k + 1LL * ways[j][i] * (j != 0);
    }
  }
  ans /= 2LL;
  cout << ans;
  puts("");
  return 0;
}

### Prompt

Please formulate a CPP solution to 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;
using namespace std;
const int K = 7;
const int N = 200500;
int n, k;
int sz[K][N], ups[K][N];
long long len[K][N], ways[K][N];
long long f[K][N], fuh[K][N], f1[K][N], fup[K][N];
vector<int> g[N];
void dfs(int ver, int prev) {
  f[0][ver] = 1;
  sz[0][ver] = 1;
  for (int i = 0; i < (int)g[ver].size(); i++) {
    int to = g[ver][i];
    if (to != prev) {
      dfs(to, ver);
      for (int j = 0; j < k; j++) {
        sz[(j + 1) % k][ver] += sz[j][to];
        f[(j + 1) % k][ver] += f[j][to];
        f1[(j + 1) % k][ver] += (f1[j][to] + 1LL * sz[j][to]);
      }
    }
  }
}
void dfs_up(int ver, int prev) {
  ++fup[0][ver];
  ++ups[0][ver];
  long long all[k];
  long long kek[k];
  int ps[k];
  for (int j = 0; j < k; ++j) ps[j] = 0;
  for (int j = 0; j < k; ++j) all[j] = kek[j] = 0LL;
  for (int i = 0; i < (int)g[ver].size(); i++) {
    int to = g[ver][i];
    if (to != prev) {
      for (int j = 0; j < k; j++) {
        all[j] += f[j][to];
        ps[j] += sz[j][to];
        kek[j] += f1[j][to];
      }
    }
  }
  for (int i = 0; i < (int)g[ver].size(); i++) {
    int to = g[ver][i];
    if (to == prev) {
      continue;
    }
    for (int j = 0; j < k; j++) {
      all[j] -= f[j][to];
      ps[j] -= sz[j][to];
      kek[j] -= f1[j][to];
    }
    for (int j = 0; j < k; j++) {
      fup[(j + 2) % k][to] += all[j];
      fup[(j + 1) % k][to] += fup[j][ver];
      fuh[(j + 2) % k][to] += (kek[j] + 2LL * ps[j]);
      fuh[(j + 1) % k][to] += (fuh[j][ver] + 1LL * ups[j][ver]);
      ups[(j + 1) % k][to] += ups[j][ver];
      ups[(j + 2) % k][to] += ps[j];
    }
    for (int j = 0; j < k; j++) {
      all[j] += f[j][to];
      ps[j] += sz[j][to];
      kek[j] += f1[j][to];
    }
  }
  for (int i = 0; i < (int)g[ver].size(); i++) {
    int to = g[ver][i];
    if (to != prev) {
      dfs_up(to, ver);
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 1);
  dfs_up(1, 1);
  long long ans = 0LL;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < k; j++) {
      ways[j][i] = f[j][i] + fup[j][i];
      len[j][i] = f1[j][i] + fuh[j][i];
      ans =
          ans + (len[j][i] - ways[j][i] * j) / k + 1LL * ways[j][i] * (j != 0);
    }
  }
  ans /= 2LL;
  cout << ans;
  puts("");
  return 0;
}

```