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;
long long N, K, dp[200100][6], cnt[200100][6], sz[200100], res;
vector<int> g[200100];
void comb(int x, int y) {
  int i;
  for (i = 1; i < K; i++) {
    dp[x][i] += dp[y][i - 1];
    cnt[x][i] += cnt[y][i - 1];
  }
  cnt[x][0] += cnt[y][K - 1];
  dp[x][0] += dp[y][K - 1];
  dp[x][1] += cnt[y][0];
}
void rmv(int x, int y) {
  int i;
  for (i = 1; i < K; i++) {
    dp[x][i] -= dp[y][i - 1];
    cnt[x][i] -= cnt[y][i - 1];
  }
  cnt[x][0] -= cnt[y][K - 1];
  dp[x][0] -= dp[y][K - 1];
  dp[x][1] -= cnt[y][0];
}
void dfs1(int nod, int f) {
  for (int x : g[nod]) {
    if (x == f) continue;
    dfs1(x, nod);
    comb(nod, x);
  }
  cnt[nod][0]++;
}
void dfs2(int nod, int f) {
  int i;
  for (i = 0; i < K; i++) res += dp[nod][i];
  for (int x : g[nod]) {
    if (x == f) continue;
    rmv(nod, x);
    comb(x, nod);
    dfs2(x, nod);
    rmv(x, nod);
    comb(nod, x);
  }
}
void dfs(int nod, int f) {
  for (int x : g[nod]) {
    if (x == f) continue;
    dfs(x, nod);
    sz[nod] += sz[x];
  }
  sz[nod]++;
  res += sz[nod] * (N - sz[nod]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> K;
  int i, x, y;
  for (i = 1; i < N; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  if (K == 1) {
    dfs(1, -1);
    cout << res << "\n";
    return 0;
  }
  dfs1(1, -1);
  dfs2(1, -1);
  cout << res / 2 << "\n";
  return 0;
}
