#include <bits/stdc++.h>
using namespace std;
const int MAXN = 4e5 + 11;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-7;
const int BUFL = 1e6 + 11;
char Obuf[BUFL], Ibuf[BUFL];
char *optr = Obuf, *ist = 0, *ied = 0;
streambuf *sb, *ssb;
inline void print(char c) {
  optr == Obuf + BUFL ? (sb->sputn(Obuf, BUFL), optr = Obuf),
      *optr++ = c     : *optr++ = c;
}
inline void flush() { sb->sputn(Obuf, optr - Obuf); }
inline char pick() {
  return ist == ied ? ied = Ibuf + ssb->sgetn(ist = Ibuf, BUFL),
                      (ist == ied ? -1 : *ist++) : *ist++;
}
template <class T>
inline void print(T x) {
  if (x == 0) {
    print(char(48));
    return;
  }
  if (x < 0) {
    print(char('-'));
    x = -x;
  }
  static char buf[233];
  register int cnt;
  for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
  while (cnt) print(buf[cnt--]);
}
template <class T>
inline int read(T &t) {
  long long x;
  register char c, f;
  f = 1;
  while (!((c = pick()) >= '0' && c <= '9') && (c ^ -1) && (c ^ 45))
    ;
  if (c == -1) return EOF;
  if (c == 45) c = pick(), f = 0;
  for (x = c - 48; (c = pick()) >= '0' && c <= '9';
       (x = ((x << 3) + (x << 1))) += c - 48)
    ;
  t = f ? x : -x;
  return 1;
}
int to[MAXN << 1], nxt[MAXN << 1], head[MAXN], tot;
long long cost[MAXN << 1];
void init(int n) {
  memset(head, -1, sizeof(head));
  tot = 0;
}
void add(int u, int v, long long w = 0) {
  to[tot] = v;
  cost[tot] = w;
  nxt[tot] = head[u];
  head[u] = tot++;
}
int n, k;
int size[MAXN], msize[MAXN][5];
inline int sub(int a, int b, int mod) {
  return ((a % mod - b % mod) % mod + mod) % mod;
}
long long dfs(int u, int fa, int d) {
  for (int i = 0; i <= 4; i++) msize[u][i] = 0;
  size[u] = 1;
  msize[u][d % k] = 1;
  long long ans = 0;
  for (int i = head[u]; ~i; i = nxt[i]) {
    int v = to[i];
    if (v == fa) continue;
    ans += dfs(v, u, d + 1);
    ans += (long long)size[v] * (n - size[v]);
    for (int i = 0; i <= k - 1; i++) {
      for (int j = 0; j <= k - 1; j++) {
        int r = sub(i + j, d << 1, k);
        int t = sub(k, r, k);
        ans += (long long)t * msize[u][i] * msize[v][j];
      }
    }
    for (int i = 0; i <= k - 1; i++) msize[u][i] += msize[v][i];
    size[u] += size[v];
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ssb = cin.rdbuf();
  sb = cout.rdbuf();
  ;
  while (~read(n)) {
    read(k);
    init(n);
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      read(u);
      read(v);
      add(u, v);
      add(v, u);
    }
    print(dfs(1, -1, 0) / k), print('\n');
    ;
  }
  flush();
  return 0;
}

### Prompt

Please formulate 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 int MAXN = 4e5 + 11;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-7;
const int BUFL = 1e6 + 11;
char Obuf[BUFL], Ibuf[BUFL];
char *optr = Obuf, *ist = 0, *ied = 0;
streambuf *sb, *ssb;
inline void print(char c) {
  optr == Obuf + BUFL ? (sb->sputn(Obuf, BUFL), optr = Obuf),
      *optr++ = c     : *optr++ = c;
}
inline void flush() { sb->sputn(Obuf, optr - Obuf); }
inline char pick() {
  return ist == ied ? ied = Ibuf + ssb->sgetn(ist = Ibuf, BUFL),
                      (ist == ied ? -1 : *ist++) : *ist++;
}
template <class T>
inline void print(T x) {
  if (x == 0) {
    print(char(48));
    return;
  }
  if (x < 0) {
    print(char('-'));
    x = -x;
  }
  static char buf[233];
  register int cnt;
  for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
  while (cnt) print(buf[cnt--]);
}
template <class T>
inline int read(T &t) {
  long long x;
  register char c, f;
  f = 1;
  while (!((c = pick()) >= '0' && c <= '9') && (c ^ -1) && (c ^ 45))
    ;
  if (c == -1) return EOF;
  if (c == 45) c = pick(), f = 0;
  for (x = c - 48; (c = pick()) >= '0' && c <= '9';
       (x = ((x << 3) + (x << 1))) += c - 48)
    ;
  t = f ? x : -x;
  return 1;
}
int to[MAXN << 1], nxt[MAXN << 1], head[MAXN], tot;
long long cost[MAXN << 1];
void init(int n) {
  memset(head, -1, sizeof(head));
  tot = 0;
}
void add(int u, int v, long long w = 0) {
  to[tot] = v;
  cost[tot] = w;
  nxt[tot] = head[u];
  head[u] = tot++;
}
int n, k;
int size[MAXN], msize[MAXN][5];
inline int sub(int a, int b, int mod) {
  return ((a % mod - b % mod) % mod + mod) % mod;
}
long long dfs(int u, int fa, int d) {
  for (int i = 0; i <= 4; i++) msize[u][i] = 0;
  size[u] = 1;
  msize[u][d % k] = 1;
  long long ans = 0;
  for (int i = head[u]; ~i; i = nxt[i]) {
    int v = to[i];
    if (v == fa) continue;
    ans += dfs(v, u, d + 1);
    ans += (long long)size[v] * (n - size[v]);
    for (int i = 0; i <= k - 1; i++) {
      for (int j = 0; j <= k - 1; j++) {
        int r = sub(i + j, d << 1, k);
        int t = sub(k, r, k);
        ans += (long long)t * msize[u][i] * msize[v][j];
      }
    }
    for (int i = 0; i <= k - 1; i++) msize[u][i] += msize[v][i];
    size[u] += size[v];
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ssb = cin.rdbuf();
  sb = cout.rdbuf();
  ;
  while (~read(n)) {
    read(k);
    init(n);
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      read(u);
      read(v);
      add(u, v);
      add(v, u);
    }
    print(dfs(1, -1, 0) / k), print('\n');
    ;
  }
  flush();
  return 0;
}

```