#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int k, n;
vector<int> adj[N];
long long sum = 0;
int cc[N][6], sz[N];
long long go[N];
long long dfs(int first, int p) {
  int u, i, j;
  long long ret = 0;
  sz[first] = 1;
  for (i = 0; i < adj[first].size(); ++i) {
    u = adj[first][i];
    if (u != p) {
      ret += dfs(u, first);
      sz[first] += sz[u];
      for (j = 0; j <= k; ++j) cc[first][j] += cc[u][j];
    }
  }
  u = cc[first][k];
  for (i = k - 1; i >= 0; --i) {
    cc[first][i + 1] = cc[first][i];
  }
  cc[first][1] += u;
  ++cc[first][1];
  go[first] = ret + u + 1;
  sum += ret;
  return go[first];
}
void dfs2(int first, int p, int *bit, long long tot) {
  int u, i, j, ori[6];
  for (i = 0; i <= k; ++i) ori[i] = bit[i];
  u = bit[k];
  for (i = k - 1; i >= 0; --i) bit[i + 1] = bit[i];
  bit[1] += u;
  tot += u;
  sum += tot;
  for (i = 0; i < adj[first].size(); ++i) {
    u = adj[first][i];
    if (u != p) {
      tot += go[u];
      for (j = 0; j <= k; ++j) {
        bit[j] += cc[u][j];
      }
    }
  }
  for (i = 0; i < adj[first].size(); ++i) {
    u = adj[first][i];
    if (u != p) {
      tot -= go[u];
      for (j = 0; j <= k; ++j) {
        bit[j] -= cc[u][j];
      }
      ++bit[k];
      dfs2(u, first, bit, tot);
      --bit[k];
      tot += go[u];
      for (j = 0; j <= k; ++j) {
        bit[j] += cc[u][j];
      }
    }
  }
  for (i = 0; i <= k; ++i) bit[i] = ori[i];
}
int main() {
  int i, a, b;
  cin >> n >> k;
  for (i = 0; i < n - 1; ++i) {
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0);
  int bit[6] = {0, 0, 0, 0, 0, 0};
  dfs2(1, 0, bit, 0);
  cout << (sum + 1) / 2 << endl;
}

### 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 N = 2e5 + 5;
int k, n;
vector<int> adj[N];
long long sum = 0;
int cc[N][6], sz[N];
long long go[N];
long long dfs(int first, int p) {
  int u, i, j;
  long long ret = 0;
  sz[first] = 1;
  for (i = 0; i < adj[first].size(); ++i) {
    u = adj[first][i];
    if (u != p) {
      ret += dfs(u, first);
      sz[first] += sz[u];
      for (j = 0; j <= k; ++j) cc[first][j] += cc[u][j];
    }
  }
  u = cc[first][k];
  for (i = k - 1; i >= 0; --i) {
    cc[first][i + 1] = cc[first][i];
  }
  cc[first][1] += u;
  ++cc[first][1];
  go[first] = ret + u + 1;
  sum += ret;
  return go[first];
}
void dfs2(int first, int p, int *bit, long long tot) {
  int u, i, j, ori[6];
  for (i = 0; i <= k; ++i) ori[i] = bit[i];
  u = bit[k];
  for (i = k - 1; i >= 0; --i) bit[i + 1] = bit[i];
  bit[1] += u;
  tot += u;
  sum += tot;
  for (i = 0; i < adj[first].size(); ++i) {
    u = adj[first][i];
    if (u != p) {
      tot += go[u];
      for (j = 0; j <= k; ++j) {
        bit[j] += cc[u][j];
      }
    }
  }
  for (i = 0; i < adj[first].size(); ++i) {
    u = adj[first][i];
    if (u != p) {
      tot -= go[u];
      for (j = 0; j <= k; ++j) {
        bit[j] -= cc[u][j];
      }
      ++bit[k];
      dfs2(u, first, bit, tot);
      --bit[k];
      tot += go[u];
      for (j = 0; j <= k; ++j) {
        bit[j] += cc[u][j];
      }
    }
  }
  for (i = 0; i <= k; ++i) bit[i] = ori[i];
}
int main() {
  int i, a, b;
  cin >> n >> k;
  for (i = 0; i < n - 1; ++i) {
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0);
  int bit[6] = {0, 0, 0, 0, 0, 0};
  dfs2(1, 0, bit, 0);
  cout << (sum + 1) / 2 << endl;
}

```