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 mod = 1e9 + 7;
const int maxn = 2e5 + 5;
const int INF = 0x7fffffff;
const long long inf = 0x3f3f3f3f;
const double eps = 1e-8;
int n, k;
vector<int> v[maxn];
struct node {
  long long k[6];
  long long num, ct;
} dp[maxn];
int du[maxn], used[maxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int u, w;
    scanf("%d%d", &u, &w);
    v[u].push_back(w);
    v[w].push_back(u);
    du[u]++;
    du[w]++;
  }
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (du[i] == 1) {
      q.push(i);
    }
    dp[i].k[0] = 1;
    dp[i].num = 0;
    dp[i].ct = 1;
  }
  long long ans = 0;
  while (q.size()) {
    int u = q.front();
    q.pop();
    if (used[u]) continue;
    used[u] = 1;
    for (auto i : v[u]) {
      if (used[i] == 1)
        continue;
      else {
        ans += dp[u].num * dp[i].ct + dp[i].num * dp[u].ct;
        ans += dp[u].k[0] * dp[i].k[0];
        for (int j = 0; j < k; j++) {
          for (int l = 0; l < k; l++) {
            if (j + l >= k + 1) {
              ans -= dp[i].k[j] * dp[u].k[l];
            }
          }
        }
        for (int j = 0; j < k; j++) {
          dp[i].k[j] += dp[u].k[(j + 1) % k];
        }
        dp[i].num += dp[u].num + dp[u].k[0];
        dp[i].ct += dp[u].ct;
        du[i]--;
        if (du[i] <= 1 && !used[i]) {
          q.push(i);
        }
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}
