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>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
int n, k;
vector<int> neighbors, a, b, sver;
vector<vector<int>> neighborhood;
vector<vector<long long>> pairss;
vector<long long> ssum, sdis;
vector<vector<int>> ver;
vector<bool> nvis;
void dfs(const int& v) {
  int prnt = -1;
  nvis[v] = false;
  pairss[v].resize(k);
  ver[v].resize(k);
  sver[v] = 0;
  sdis[v] = 0;
  ssum[v] = 0;
  for (int j = 0; j < k; ++j) {
    ver[v][j] = 0;
    pairss[v][j] = 0;
  }
  for (int i = 0; i < neighbors[v]; ++i) {
    if (nvis[neighborhood[v][i]]) {
      dfs(neighborhood[v][i]);
      sver[v] += sver[neighborhood[v][i]];
      for (int j = 0; j < k; ++j) {
        ver[v][(j + 1) % k] += ver[neighborhood[v][i]][j];
        pairss[v][j] += pairss[neighborhood[v][i]][j];
        pairss[v][(j + 1) % k] += ver[neighborhood[v][i]][j];
      }
      sdis[v] += sdis[neighborhood[v][i]] + sver[neighborhood[v][i]];
      ssum[v] += ssum[neighborhood[v][i]] + sdis[neighborhood[v][i]] +
                 sver[neighborhood[v][i]];
    } else
      prnt = neighborhood[v][i];
  }
  vector<long long> adding(k);
  for (int j = 0; j < k; ++j) adding[j] = 0;
  for (int i = 0; i < neighbors[v]; ++i) {
    if (neighborhood[v][i] != prnt) {
      ssum[v] += (sdis[neighborhood[v][i]] + sver[neighborhood[v][i]]) *
                 (sver[v] - sver[neighborhood[v][i]]);
      for (int j = 0; j < k; ++j) {
        for (int jj = 0; jj < k; ++jj) {
          adding[(j + jj + 2) % k] +=
              static_cast<long long>(ver[neighborhood[v][i]][j]) *
              (ver[v][(jj + 1) % k] - ver[neighborhood[v][i]][jj]);
        }
      }
    }
  }
  for (int j = 0; j < k; ++j) pairss[v][j] += adding[j] / 2;
  ++ver[v][0];
  ++sver[v];
}
int main() {
  cin >> n >> k;
  a.resize(n);
  b.resize(n);
  neighbors.resize(n);
  nvis.resize(n);
  pairss.resize(n);
  ver.resize(n);
  sver.resize(n);
  ssum.resize(n);
  sdis.resize(n);
  for (int i = 0; i < n; ++i) {
    neighbors[i] = 0;
    nvis[i] = true;
  }
  for (int i = 1; i < n; ++i) {
    cin >> a[i] >> b[i];
    --a[i];
    --b[i];
    ++neighbors[a[i]];
    ++neighbors[b[i]];
  }
  neighborhood.resize(n);
  for (int i = 0; i < n; ++i) {
    neighborhood[i].resize(neighbors[i]);
    neighbors[i] = 0;
  }
  for (int i = 1; i < n; ++i) {
    neighborhood[a[i]][neighbors[a[i]]++] = b[i];
    neighborhood[b[i]][neighbors[b[i]]++] = a[i];
  }
  dfs(0);
  long long ans = ssum[0];
  for (int i = 1; i < k; ++i) ans += (k - i) * pairss[0][i];
  cout << ans / k << endl;
  return 0;
}
