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;
vector<long long> g[200002];
vector<pair<int, int> > edges;
int p[200002];
long long snode[200002];
long long plsum = 0;
int n, k;
void dfs(int s) {
  snode[s] = 1;
  for (int i = 0; i < g[s].size(); i++) {
    int v = g[s][i];
    if (v != p[s]) {
      p[v] = s;
      dfs(v);
      snode[s] = snode[s] + snode[v];
    }
  }
}
void cmp_pathlensum() {
  for (int i = 0; i < edges.size(); i++) {
    int a, b;
    a = edges[i].first;
    b = edges[i].second;
    if (b == p[a]) swap(a, b);
    plsum = plsum + snode[b] * (n - snode[b]);
  }
}
long long dp[200002][2][5];
void cmp_rem(int s) {
  long long child = 0;
  long long ch_rem[5] = {0};
  long long ch_rem2[5] = {0};
  for (int i = 0; i < g[s].size(); i++) {
    int v = g[s][i];
    if (v != p[s]) {
      cmp_rem(v);
      child++;
      for (int r = 0; r < k; r++) {
        ch_rem[r] += dp[v][0][r];
        ch_rem2[r] += dp[v][1][r];
      }
    }
  }
  for (int r = 0; r < k; r++) {
    dp[s][0][r] = ch_rem[(r - 1 + k) % k];
  }
  dp[s][0][1] += child;
  long long sum[5] = {0};
  if (child >= 2) {
    for (int r = 0; r < k; r++) {
      for (int i = 0; i < g[s].size(); i++) {
        int v = g[s][i];
        if (v != p[s]) {
          for (int rem = 0; rem < k; rem++) {
            int rv = ((r - rem - 2) % k + k) % k;
            sum[r] += dp[v][0][rem] * (ch_rem[rv] - dp[v][0][rv]);
          }
        }
      }
      sum[r] = sum[r] / 2;
    }
    for (int r = 0; r < k; r++) {
      for (int i = 0; i < g[s].size(); i++) {
        int v = g[s][i];
        if (v != p[s]) {
          int rv = ((r - 2) % k + k) % k;
          sum[r] = sum[r] + (ch_rem[rv] - dp[v][0][rv]);
        }
      }
    }
    if (k == 2) {
      sum[0] = sum[0] + (child - 1) * child / 2;
    } else if (k > 2) {
      sum[2] = sum[2] + (child - 1) * child / 2;
    }
  }
  for (int r = 0; r < k; r++) {
    dp[s][1][r] = dp[s][0][r] + ch_rem2[r] + sum[r];
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    edges.push_back(make_pair(a, b));
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1);
  cmp_pathlensum();
  if (k == 1) {
    cout << plsum << endl;
    return 0;
  }
  cmp_rem(1);
  long long sum = plsum;
  for (int r = 0; r < k; r++) {
    sum = sum + ((k - r) % k) * dp[1][1][r];
  }
  long long ans = sum / k;
  cout << ans << endl;
  return 0;
}
