#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T LCM(T a, T b) {
  return a * (b / gcd(a, b));
}
template <typename T>
T expo(T base, T e, T mod) {
  T res = 1;
  while (e > 0) {
    if (e & 1) res = res * base % mod;
    base = base * base % mod;
    e >>= 1;
  }
  return res;
}
template <typename T, typename second>
T expo(T b, second e) {
  if (e <= 1) return e == 0 ? 1 : b;
  return (e & 1) == 0 ? expo((b * b), e >> 1) : (b * expo((b * b), e >> 1));
}
template <typename T, typename second>
T modinv(T a, second mod) {
  return expo(a, mod - 2, mod);
}
template <class T, class second>
std::ostream &operator<<(std::ostream &os, const std::pair<T, second> &t) {
  os << "(" << t.first << ", " << t.second << ")";
  return os;
}
template <class T>
std::ostream &operator<<(std::ostream &os, const std::vector<T> &t) {
  os << "[";
  for (__typeof(t.begin()) it = t.begin(); it != t.end(); it++) {
    if (it != t.begin()) os << ", ";
    os << *it;
  }
  os << "]";
  return os;
}
template <typename T>
void in(T &x) {
  T c = getchar();
  while (((c < 48) || (c > 57)) && (c != '-')) c = getchar();
  bool neg = false;
  if (c == '-') neg = true;
  x = 0;
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) x = (x * 10) + (c - 48);
  if (neg) x = -x;
}
const int MAXN = 2e5 + 3;
int n, k;
vector<int> adj[MAXN];
int cnt_down[MAXN][5];
int cnt_up[MAXN][5];
long long sum = 0, res = 0;
void compute_down(int r, int p) {
  for (auto &elem : adj[r]) {
    if (elem != p) {
      compute_down(elem, r);
      for (int i = 0; i < k; i++) {
        cnt_down[r][(i + 1) % k] += cnt_down[elem][i];
      }
    }
  }
  cnt_down[r][0]++;
}
void compute_up(int r, int p) {
  int temp[5] = {0};
  for (auto &elem : adj[r]) {
    if (elem != p) {
      for (int i = 0; i < k; i++) {
        temp[i] += cnt_down[elem][i];
      }
    }
  }
  for (auto &elem : adj[r]) {
    if (elem != p) {
      for (int i = 0; i < k; i++) {
        cnt_up[elem][(i + 1) % k] += cnt_up[r][i];
        temp[i] -= cnt_down[elem][i];
      }
      for (int i = 0; i < k; i++) {
        cnt_up[elem][(i + 2) % k] += temp[i];
      }
      cnt_up[elem][1 % k]++;
      for (int i = 0; i < k; i++) temp[i] += cnt_down[elem][i];
      compute_up(elem, r);
    }
  }
}
void dfs(int r, int p, int d) {
  sum += (d + k - 1) / k;
  for (auto &elem : adj[r]) {
    if (elem != p) {
      dfs(elem, r, d + 1);
    }
  }
}
void go(int r, int p, long long sum) {
  res += sum;
  for (auto &elem : adj[r]) {
    if (elem != p) {
      long long nsum = sum + cnt_up[elem][1 % k] - cnt_down[elem][0];
      go(elem, r, nsum);
    }
  }
}
int main() {
  int a, b;
  in(n), in(k);
  for (int i = 1; i < n; i++) {
    in(a), in(b);
    adj[a].push_back(b), adj[b].push_back(a);
  }
  compute_down(1, 0);
  compute_up(1, 0);
  dfs(1, 0, 0);
  go(1, 0, sum);
  printf("%I64d\n", res / 2);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 mod = 1e9 + 7;
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T LCM(T a, T b) {
  return a * (b / gcd(a, b));
}
template <typename T>
T expo(T base, T e, T mod) {
  T res = 1;
  while (e > 0) {
    if (e & 1) res = res * base % mod;
    base = base * base % mod;
    e >>= 1;
  }
  return res;
}
template <typename T, typename second>
T expo(T b, second e) {
  if (e <= 1) return e == 0 ? 1 : b;
  return (e & 1) == 0 ? expo((b * b), e >> 1) : (b * expo((b * b), e >> 1));
}
template <typename T, typename second>
T modinv(T a, second mod) {
  return expo(a, mod - 2, mod);
}
template <class T, class second>
std::ostream &operator<<(std::ostream &os, const std::pair<T, second> &t) {
  os << "(" << t.first << ", " << t.second << ")";
  return os;
}
template <class T>
std::ostream &operator<<(std::ostream &os, const std::vector<T> &t) {
  os << "[";
  for (__typeof(t.begin()) it = t.begin(); it != t.end(); it++) {
    if (it != t.begin()) os << ", ";
    os << *it;
  }
  os << "]";
  return os;
}
template <typename T>
void in(T &x) {
  T c = getchar();
  while (((c < 48) || (c > 57)) && (c != '-')) c = getchar();
  bool neg = false;
  if (c == '-') neg = true;
  x = 0;
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) x = (x * 10) + (c - 48);
  if (neg) x = -x;
}
const int MAXN = 2e5 + 3;
int n, k;
vector<int> adj[MAXN];
int cnt_down[MAXN][5];
int cnt_up[MAXN][5];
long long sum = 0, res = 0;
void compute_down(int r, int p) {
  for (auto &elem : adj[r]) {
    if (elem != p) {
      compute_down(elem, r);
      for (int i = 0; i < k; i++) {
        cnt_down[r][(i + 1) % k] += cnt_down[elem][i];
      }
    }
  }
  cnt_down[r][0]++;
}
void compute_up(int r, int p) {
  int temp[5] = {0};
  for (auto &elem : adj[r]) {
    if (elem != p) {
      for (int i = 0; i < k; i++) {
        temp[i] += cnt_down[elem][i];
      }
    }
  }
  for (auto &elem : adj[r]) {
    if (elem != p) {
      for (int i = 0; i < k; i++) {
        cnt_up[elem][(i + 1) % k] += cnt_up[r][i];
        temp[i] -= cnt_down[elem][i];
      }
      for (int i = 0; i < k; i++) {
        cnt_up[elem][(i + 2) % k] += temp[i];
      }
      cnt_up[elem][1 % k]++;
      for (int i = 0; i < k; i++) temp[i] += cnt_down[elem][i];
      compute_up(elem, r);
    }
  }
}
void dfs(int r, int p, int d) {
  sum += (d + k - 1) / k;
  for (auto &elem : adj[r]) {
    if (elem != p) {
      dfs(elem, r, d + 1);
    }
  }
}
void go(int r, int p, long long sum) {
  res += sum;
  for (auto &elem : adj[r]) {
    if (elem != p) {
      long long nsum = sum + cnt_up[elem][1 % k] - cnt_down[elem][0];
      go(elem, r, nsum);
    }
  }
}
int main() {
  int a, b;
  in(n), in(k);
  for (int i = 1; i < n; i++) {
    in(a), in(b);
    adj[a].push_back(b), adj[b].push_back(a);
  }
  compute_down(1, 0);
  compute_up(1, 0);
  dfs(1, 0, 0);
  go(1, 0, sum);
  printf("%I64d\n", res / 2);
  return 0;
}

```