#include <bits/stdc++.h>
using namespace std;
const int INF = 0x7fffffff, SINF = 0x3f3f3f3f;
const double PI = 3.141592653589793;
const long long LINF = 0x7fffffffffffffffLL, SLINF = 0x3f3f3f3f3f3f3f3fLL;
int _v, _f, _c;
long long _l;
void read() {
  _c = getchar(), _v = _f = 0;
  while (_c < '0' || _c > '9') {
    if (_c == -1) {
      _f = -1;
      return;
    } else if (_c == '-') {
      _f = 1;
    }
    _c = getchar();
  }
  while (_c <= '9' && _c >= '0') {
    _v = (_v << 3) + (_v << 1) + (_c ^ 48);
    _c = getchar();
  }
  if (_f) {
    _v = -_v;
  }
}
void read_ll() {
  _c = getchar(), _l = _f = 0;
  while (_c < '0' || _c > '9') {
    if (_c == -1) {
      _f = -1;
      return;
    } else if (_c == '-') {
      _f = 1;
    }
    _c = getchar();
  }
  while (_c <= '9' && _c >= '0') {
    _l = (_l << 3) + (_l << 1) + (_c ^ 48);
    _c = getchar();
  }
  if (_f) {
    _l = -_l;
  }
}
void read_str(char* _s) {
  _v = 0, _c = getchar();
  while (_c != '\n' && _c != ' ' && _c != -1) {
    _s[_v++] = (char)_c;
    _c = getchar();
  }
  if (_c == -1) _f = -1;
}
const int MAXN = 200005, MAXM = 100005, MAXK = 100005, MOD = 1e9 + 7;
const double EPS = 1e-8;
int n, k;
struct Edge {
  int t, n;
} edge[MAXN * 2];
int head[MAXN], en;
void add_e(int f, int t) {
  edge[en].t = t, edge[en].n = head[f], head[f] = en++;
}
int siz[MAXN];
long long dp[MAXN][5], dp2[MAXN][5], tot[5];
void dp_work(int cur = 0, int last = -1) {
  siz[cur] = 1;
  for (int i = head[cur]; ~i; i = edge[i].n) {
    if (edge[i].t != last) {
      dp_work(edge[i].t, cur);
      siz[cur] += siz[edge[i].t];
      for (int j = 0; j < k; j++) {
        for (int l = 0; l < k; l++) {
          dp[cur][(j + l + 1) % k] += dp2[cur][j] * dp2[edge[i].t][l];
        }
        dp[cur][(j + 1) % k] += dp2[cur][j];
      }
      dp2[cur][1 % k]++;
      for (int j = 0; j < k; j++) {
        dp2[cur][(j + 1) % k] += dp2[edge[i].t][j];
      }
    }
  }
  for (int i = 0; i < k; i++) {
    tot[i] += dp[cur][i] + dp2[cur][i];
  }
}
int main() {
  memset(head, -1, sizeof(head));
  n = (read(), _v), k = (read(), _v);
  for (int i = 1; i < n; i++) {
    int f = (read(), _v) - 1, t = (read(), _v) - 1;
    add_e(f, t), add_e(t, f);
  }
  dp_work();
  long long flen = 0;
  for (int i = 1; i < n; i++) {
    flen += (long long)(siz[i]) * (n - siz[i]);
  }
  for (int i = 1; i < k; i++) {
    flen += tot[i] * (k - i);
  }
  printf("%lld\n", flen / k);
  return 0;
}

### Prompt

Generate 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 INF = 0x7fffffff, SINF = 0x3f3f3f3f;
const double PI = 3.141592653589793;
const long long LINF = 0x7fffffffffffffffLL, SLINF = 0x3f3f3f3f3f3f3f3fLL;
int _v, _f, _c;
long long _l;
void read() {
  _c = getchar(), _v = _f = 0;
  while (_c < '0' || _c > '9') {
    if (_c == -1) {
      _f = -1;
      return;
    } else if (_c == '-') {
      _f = 1;
    }
    _c = getchar();
  }
  while (_c <= '9' && _c >= '0') {
    _v = (_v << 3) + (_v << 1) + (_c ^ 48);
    _c = getchar();
  }
  if (_f) {
    _v = -_v;
  }
}
void read_ll() {
  _c = getchar(), _l = _f = 0;
  while (_c < '0' || _c > '9') {
    if (_c == -1) {
      _f = -1;
      return;
    } else if (_c == '-') {
      _f = 1;
    }
    _c = getchar();
  }
  while (_c <= '9' && _c >= '0') {
    _l = (_l << 3) + (_l << 1) + (_c ^ 48);
    _c = getchar();
  }
  if (_f) {
    _l = -_l;
  }
}
void read_str(char* _s) {
  _v = 0, _c = getchar();
  while (_c != '\n' && _c != ' ' && _c != -1) {
    _s[_v++] = (char)_c;
    _c = getchar();
  }
  if (_c == -1) _f = -1;
}
const int MAXN = 200005, MAXM = 100005, MAXK = 100005, MOD = 1e9 + 7;
const double EPS = 1e-8;
int n, k;
struct Edge {
  int t, n;
} edge[MAXN * 2];
int head[MAXN], en;
void add_e(int f, int t) {
  edge[en].t = t, edge[en].n = head[f], head[f] = en++;
}
int siz[MAXN];
long long dp[MAXN][5], dp2[MAXN][5], tot[5];
void dp_work(int cur = 0, int last = -1) {
  siz[cur] = 1;
  for (int i = head[cur]; ~i; i = edge[i].n) {
    if (edge[i].t != last) {
      dp_work(edge[i].t, cur);
      siz[cur] += siz[edge[i].t];
      for (int j = 0; j < k; j++) {
        for (int l = 0; l < k; l++) {
          dp[cur][(j + l + 1) % k] += dp2[cur][j] * dp2[edge[i].t][l];
        }
        dp[cur][(j + 1) % k] += dp2[cur][j];
      }
      dp2[cur][1 % k]++;
      for (int j = 0; j < k; j++) {
        dp2[cur][(j + 1) % k] += dp2[edge[i].t][j];
      }
    }
  }
  for (int i = 0; i < k; i++) {
    tot[i] += dp[cur][i] + dp2[cur][i];
  }
}
int main() {
  memset(head, -1, sizeof(head));
  n = (read(), _v), k = (read(), _v);
  for (int i = 1; i < n; i++) {
    int f = (read(), _v) - 1, t = (read(), _v) - 1;
    add_e(f, t), add_e(t, f);
  }
  dp_work();
  long long flen = 0;
  for (int i = 1; i < n; i++) {
    flen += (long long)(siz[i]) * (n - siz[i]);
  }
  for (int i = 1; i < k; i++) {
    flen += tot[i] * (k - i);
  }
  printf("%lld\n", flen / k);
  return 0;
}

```