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 N = 200005;
bool us[N];
long long n, k, i, j, x, y, ans, kol[6][N], d[N], a[6], b[6], kk[N];
vector<long long> g[N];
void dfs(int x) {
  us[x] = 1;
  long long y;
  for (long long i = 0; i < g[x].size(); ++i) {
    y = g[x][i];
    if (us[y]) continue;
    dfs(y);
    kol[1][x]++;
    d[x]++;
    d[x] += d[y];
    for (long long j = 1; j < k; ++j) {
      kol[j + 1][x] += kol[j][y];
    }
    kol[1][x] += kol[k][y];
    d[x] += kol[k][y];
  }
}
void dfs2(long long x, long long s) {
  us[x] = 1;
  ans += d[x];
  ans += s;
  long long b[6] = {0}, sum[6] = {0};
  for (long long i = 1; i <= k; ++i) {
    b[i] = a[i];
  }
  long long y;
  for (long long i = 0; i < g[x].size(); ++i) {
    y = g[x][i];
    if (us[y]) continue;
    for (long long j = 2; j <= k; ++j) {
      sum[j] = kol[j][x] - kol[j - 1][y] + b[j];
    }
    sum[1] = kol[1][x] - kol[k][y] + b[1];
    sum[1]--;
    for (long long j = 1; j < k; ++j) {
      a[j + 1] = sum[j];
    }
    a[1] = sum[k];
    a[1]++;
    long long z = k - 1;
    if (z == 0) z = k;
    dfs2(y, s + b[k] + d[x] - d[y] - kol[k][y] - 1 + kol[k][x] - kol[z][y] + 1);
  }
}
void dfs11(long long x) {
  long long i, y;
  us[x] = 1;
  for (i = 0; i < g[x].size(); ++i) {
    y = g[x][i];
    if (us[y]) continue;
    dfs11(y);
    d[x] += d[y];
    kk[x] += kk[y];
  }
  d[x] += kk[x];
  kk[x]++;
}
void dfs12(long long x, long long s, long long t) {
  long long i, y;
  us[x] = 1;
  ans += s;
  ans += d[x];
  for (i = 0; i < g[x].size(); ++i) {
    y = g[x][i];
    if (us[y]) continue;
    dfs12(y, s + t + kk[x] - kk[y] + d[x] - d[y] - kk[y], t + kk[x] - kk[y]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (i = 1; i < n; ++i) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  if (k == 1) {
    dfs11(1);
    for (i = 1; i <= n; ++i) {
      us[i] = 0;
    }
    dfs12(1, 0, 0);
    cout << ans / 2;
    return 0;
  }
  dfs(1);
  for (i = 1; i <= n; ++i) {
    us[i] = 0;
  }
  dfs2(1, 0);
  cout << ans / 2;
}
