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 long long MaxN = 2e5 + 100;
const long long LogN = 20;
const long long Inf = 1e18;
double eps = 1e-3;
const long long md = 1e9 + 7;
long long N, K;
long long Ans = 0;
vector<long long> Gr[MaxN];
long long dp[MaxN][5];
long long Sz[MaxN];
void dfs(long long v, long long p = -1, long long h = 0) {
  Sz[v] = dp[v][h % K] = 1;
  for (long long to : Gr[v]) {
    if (to == p) continue;
    dfs(to, v, h + 1);
    for (int i = 0; i < K; i++) {
      for (int j = 0; j < K; j++) {
        long long a = ((i + j) - 2 * h + K) % K;
        long long b = ((K)-a + K) % K;
        Ans += b * (dp[v][i]) * (dp[to][j]);
      }
    }
    for (int i = 0; i < K; i++) {
      dp[v][i] += dp[to][i];
    }
    Sz[v] += Sz[to];
  }
  Ans += Sz[v] * (N - Sz[v]);
}
int main() {
  cin >> N >> K;
  for (int i = 0; i < N - 1; i++) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    Gr[a].push_back(b);
    Gr[b].push_back(a);
  }
  dfs(0);
  cout << Ans / K;
  return 0;
}
