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 n, k;
long long int an, dp[200005][6], hand[200005];
vector<int> vt[200005];
void dfs(int node, int par) {
  for (int i = 0, tmp; i < vt[node].size(); i++) {
    tmp = vt[node][i];
    if (tmp == par) continue;
    dfs(tmp, node);
    hand[node] += (dp[tmp][k] + hand[tmp]);
    for (int j = 2; j <= k; j++) dp[node][j] += dp[tmp][j - 1];
    dp[node][1] += (dp[tmp][k] + 1);
  }
}
void dfs(int node, int par, long long int up[], long long int lol) {
  for (int i = 1; i <= k; i++) an += up[i] + dp[node][i];
  long long int a[10], tpp[10];
  an += (lol + hand[node]);
  for (int i = 0, tmp; i < vt[node].size(); i++) {
    tmp = vt[node][i];
    if (tmp == par) continue;
    for (int j = 2; j <= k; j++) tpp[j] = up[j] + dp[node][j] - dp[tmp][j - 1];
    tpp[1] = up[1] + dp[node][1] - dp[tmp][k] - 1;
    for (int j = 2; j <= k; j++) a[j] = tpp[j - 1];
    a[1] = tpp[k] + 1;
    dfs(tmp, node, a, lol + hand[node] - hand[tmp] + a[1] - 1 - dp[tmp][k]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  scanf("%d%d", &n, &k);
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    vt[x].push_back(y);
    vt[y].push_back(x);
  }
  dfs(1, 0);
  long long int up[10];
  memset(up, 0, sizeof(up));
  dfs(1, 0, up, 0);
  an >>= 1;
  printf("%lld\n", an);
  return 0;
}
