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 double EPSILON = 1e-9;
const double PI = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 0x3c3c3c3c;
const long long INFL = 0x3c3c3c3c3c3c3c3c;
const int MAX_N = 102;
int N, M, a, b;
vector<int> G[200003];
long long size[200003];
long long dp[200003][5];
long long ans;
void dfs(int here, int dad) {
  size[here] = 0;
  dp[here][0]++;
  for (int there : G[here]) {
    if (there == dad) continue;
    dfs(there, here);
    for (int i = 0; i < M; ++i) {
      for (int j = 0; j < M; ++j) {
        int k = i + j + 1;
        ans += (M - ((k % M) ? (k % M) : M)) * dp[here][i] * dp[there][j];
      }
    }
    for (int i = 0; i < M; ++i) dp[here][(i + 1) % M] += dp[there][i];
    ans += size[there] * (N - size[there]);
    size[here] += size[there];
  }
  size[here]++;
}
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 0; i < N - 1; ++i) {
    scanf("%d%d", &a, &b);
    a--, b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(0, -1);
  printf("%lld", ans / M);
  return 0;
}
