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.
#include <bits/stdc++.h>
using namespace std;
const int N = 200050;
long long Paths[5][2];
void Set(int x, int y) {
  Paths[x][0] += y;
  Paths[x][1]++;
}
vector<int> Tree[N];
int NodeSize[N];
bool vis[N];
void DFS(int cur, int prev, int &n) {
  n++;
  NodeSize[cur] = 1;
  int i;
  for (i = 0; i < Tree[cur].size(); i++) {
    int v = Tree[cur][i];
    if (v != prev && !vis[v]) {
      DFS(v, cur, n);
      NodeSize[cur] += NodeSize[v];
    }
  }
}
int Find(int cur, int prev, int n) {
  int i;
  for (i = 0; i < Tree[cur].size(); i++) {
    int v = Tree[cur][i];
    if (v != prev && !vis[v]) {
      if (NodeSize[v] > n / 2) return Find(v, cur, n);
    }
  }
  return cur;
}
int Find(int src) {
  int n = 0;
  DFS(src, src, n);
  return Find(src, src, n);
}
long long sol;
int k;
void UpdateSol(int j, int l) {
  int i;
  for (i = 0; i < k; i++) {
    sol += Paths[i][0];
    sol += Paths[i][1] * l;
    if (j + i == 0)
      sol += 0;
    else if (j + i <= k)
      sol += Paths[i][1];
    else
      sol += Paths[i][1] * 2;
  }
}
vector<int> back;
void Solve(int cur, int prev, int path) {
  UpdateSol(path % k, path / k);
  back.push_back(path);
  int i;
  for (i = 0; i < Tree[cur].size(); i++) {
    int v = Tree[cur][i];
    if (v != prev && !vis[v]) {
      Solve(v, cur, path + 1);
    }
  }
}
void Decompose(int src) {
  src = Find(src);
  vis[src] = true;
  int i, j;
  for (i = 0; i < k; i++) Paths[i][0] = 0, Paths[i][1] = 0;
  Set(0, 0);
  for (i = 0; i < Tree[src].size(); i++) {
    int v = Tree[src][i];
    if (!vis[v]) {
      Solve(v, src, 1);
      for (j = 0; j < back.size(); j++) Set(back[j] % k, back[j] / k);
      back.clear();
    }
  }
  for (i = 0; i < Tree[src].size(); i++) {
    int v = Tree[src][i];
    if (!vis[v]) Decompose(v);
  }
}
int main() {
  int n, i, u, v;
  scanf("%i %i", &n, &k);
  for (i = 1; i < n; i++) {
    scanf("%i %i", &u, &v);
    Tree[u].push_back(v);
    Tree[v].push_back(u);
  }
  Decompose(1);
  printf("%lld\n", sol);
  return 0;
}
