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;
int k;
struct Edge;
vector<Edge *> m[200000];
struct Vertex {
  long long s;
  int cnt[5];
};
Vertex v[200000];
struct Edge {
  int to;
  long long s;
  int cnt[5];
  Edge *rev;
  Edge(int to) : to(to), s(-1) {
    for (int i = 0; i < k; i++) cnt[i] = 0;
  }
  void rec() {
    int from = rev->to;
    if (rev->s != -1 && v[to].s == 0) {
      for (Edge *y : m[to]) {
        if (y->s == -1) y->rec();
        for (int i = 0; i < k; i++) v[to].cnt[i] += y->cnt[i];
        v[to].s += y->s;
      }
    }
    if (v[to].s) {
      for (int i = 0; i < k; i++) cnt[i] += v[to].cnt[i] - rev->cnt[i];
      s = v[to].s - rev->s;
    } else {
      s = 0;
      for (Edge *y : m[to]) {
        if (y->to == from) continue;
        if (y->s == -1) y->rec();
        for (int i = 0; i < k; i++) cnt[i] += y->cnt[i];
        s += y->s;
      }
    }
    int t = cnt[k - 1];
    for (int i = k - 2; i >= 0; i--) cnt[i + 1] = cnt[i];
    cnt[0] = t;
    cnt[0]++;
    s++;
    s += t;
  }
};
int main() {
  int n;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    Edge *ab = new Edge(b);
    Edge *ba = new Edge(a);
    ab->rev = ba;
    ba->rev = ab;
    m[a].push_back(ab);
    m[b].push_back(ba);
  }
  long long ss = 0;
  for (int i = 0; i < n; i++) {
    for (Edge *y : m[i]) {
      if (y->s == -1) y->rec();
      ss += y->s;
    }
  }
  cout << ss / 2;
}
