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 Set(long long n, long long pos) { return n = n | (1LL << pos); }
long long reset(long long n, long long pos) { return n = n & ~(1LL << pos); }
bool check(long long n, long long pos) { return (bool)(n & (1LL << pos)); }
long long mul(long long a, long long b) {
  long long c;
  c = (a % 1000000007 * b % 1000000007) % 1000000007;
  return c;
}
long long add(long long a, long long b) {
  long long c;
  c = (a % 1000000007 + b % 1000000007) % 1000000007;
  return c;
}
long long sub(long long a, long long b) {
  long long c;
  c = ((a % 1000000007 - b % 1000000007) % 1000000007 + 1000000007) %
      1000000007;
  return c;
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long exgcd(long long x, long long y, long long &a, long long &b) {
  int flag = 0;
  long long t, la = 1, lb = 0, ra = 0, rb = 1;
  while (x % y) {
    if (flag == 0)
      la -= x / y * ra, lb -= x / y * rb;
    else
      ra -= x / y * la, rb -= x / y * lb;
    t = x, x = y, y = t % y;
    flag = 1 - flag;
  }
  if (flag == 0)
    a = ra, b = rb;
  else
    a = la, b = lb;
  return y;
}
vector<long long> graph[200005];
long long n, k, ans;
long long subtree[200005];
long long cnt[200005][6];
void dfs(long long src, long long par) {
  subtree[src] = 1;
  for (long long i = 0; i < graph[src].size(); i++) {
    long long adj = graph[src][i];
    if (adj != par) {
      dfs(adj, src);
      subtree[src] += subtree[adj];
    }
  }
  ans += subtree[src] * (n - subtree[src]);
}
long long subtract(long long a, long long b) { return ((a - b) % k + k) % k; }
void dfs2(long long src, long long par, long long d) {
  cnt[src][d % k] = 1;
  for (long long i = 0; i < graph[src].size(); i++) {
    long long adj = graph[src][i];
    if (adj != par) {
      dfs2(adj, src, d + 1);
      for (long long j = 0; j < k; j++) {
        for (long long l = 0; l < k; l++) {
          long long rem = subtract(j + l, (2LL * d) % k);
          long long needs = subtract(k, rem);
          ans += needs * cnt[src][j] * cnt[adj][l];
        }
      }
      for (long long j = 0; j < k; j++) cnt[src][j] += cnt[adj][j];
    }
  }
}
int main() {
  scanf("%lld", &n);
  scanf("%lld", &k);
  for (long long i = 1; i <= n - 1; i++) {
    long long u, v;
    scanf("%lld", &u);
    scanf("%lld", &v);
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  dfs(1, -1);
  dfs2(1, -1, 0);
  printf("%lld\n", ans / k);
  return 0;
}
