#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5;
const long long M = 1e6 + 5;
const long long mod = 1e8 + 7;
const long long inf = 1e9;
inline long long read() {
  long long s = 0;
  register bool neg = 0;
  register char c = getchar();
  for (; c < '0' || c > '9'; c = getchar()) neg |= (c == '-');
  for (; c >= '0' && c <= '9'; s = s * 10 + (c ^ 48), c = getchar())
    ;
  s = (neg ? -s : s);
  return s;
}
long long a, b, dp[N][5], f[N];
vector<long long> st[N];
inline void dfs(long long n, long long fa) {
  for (long long i = (0); i <= (((long long)((st[n]).size())) - 1); ++i) {
    long long v = st[n][i];
    if (v == fa) continue;
    dfs(v, n);
    for (long long j = (0); j <= (b - 1); ++j)
      dp[n][j] += dp[v][(j - 1 + b) % b];
    f[1] += dp[v][b - 1];
  }
  dp[n][0]++;
}
inline void Dfs(long long n, long long fa) {
  static long long tmp[5];
  for (long long i = (0); i <= (((long long)((st[n]).size())) - 1); ++i) {
    long long v = st[n][i];
    if (v == fa) continue;
    memset(tmp, 0, sizeof(tmp));
    f[v] -= dp[v][0];
    for (long long j = (0); j <= (b - 1); ++j)
      tmp[j] = dp[n][j] - dp[v][(j - 1 + b) % b];
    f[v] += tmp[0] + f[n];
    for (long long j = (0); j <= (b - 1); ++j) dp[v][j] += tmp[(j - 1 + b) % b];
    Dfs(v, n);
  }
}
signed main() {
  a = read();
  b = read();
  long long x, y;
  for (long long i = (1); i <= (a - 1); ++i) {
    x = read();
    y = read();
    st[x].push_back(y);
    st[y].push_back(x);
  }
  dfs(1, 0);
  for (long long i = (1); i <= (b - 1); ++i) f[1] += dp[1][i];
  Dfs(1, 0);
  long long ans = 0;
  for (long long i = (1); i <= (a); ++i) ans += f[i];
  printf("%lld", ans / 2);
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5;
const long long M = 1e6 + 5;
const long long mod = 1e8 + 7;
const long long inf = 1e9;
inline long long read() {
  long long s = 0;
  register bool neg = 0;
  register char c = getchar();
  for (; c < '0' || c > '9'; c = getchar()) neg |= (c == '-');
  for (; c >= '0' && c <= '9'; s = s * 10 + (c ^ 48), c = getchar())
    ;
  s = (neg ? -s : s);
  return s;
}
long long a, b, dp[N][5], f[N];
vector<long long> st[N];
inline void dfs(long long n, long long fa) {
  for (long long i = (0); i <= (((long long)((st[n]).size())) - 1); ++i) {
    long long v = st[n][i];
    if (v == fa) continue;
    dfs(v, n);
    for (long long j = (0); j <= (b - 1); ++j)
      dp[n][j] += dp[v][(j - 1 + b) % b];
    f[1] += dp[v][b - 1];
  }
  dp[n][0]++;
}
inline void Dfs(long long n, long long fa) {
  static long long tmp[5];
  for (long long i = (0); i <= (((long long)((st[n]).size())) - 1); ++i) {
    long long v = st[n][i];
    if (v == fa) continue;
    memset(tmp, 0, sizeof(tmp));
    f[v] -= dp[v][0];
    for (long long j = (0); j <= (b - 1); ++j)
      tmp[j] = dp[n][j] - dp[v][(j - 1 + b) % b];
    f[v] += tmp[0] + f[n];
    for (long long j = (0); j <= (b - 1); ++j) dp[v][j] += tmp[(j - 1 + b) % b];
    Dfs(v, n);
  }
}
signed main() {
  a = read();
  b = read();
  long long x, y;
  for (long long i = (1); i <= (a - 1); ++i) {
    x = read();
    y = read();
    st[x].push_back(y);
    st[y].push_back(x);
  }
  dfs(1, 0);
  for (long long i = (1); i <= (b - 1); ++i) f[1] += dp[1][i];
  Dfs(1, 0);
  long long ans = 0;
  for (long long i = (1); i <= (a); ++i) ans += f[i];
  printf("%lld", ans / 2);
  return 0;
}

```