#include <bits/stdc++.h>
using namespace std;
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T1, class T2>
T1 chmin(T1& x, const T2& y) {
  if (T1(y) < x) return x = y;
  return x;
}
template <class T1, class T2>
T1 chmax(T1& x, const T2& y) {
  if (T1(y) > x) return x = y;
  return x;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << '{';
  bool was = false;
  for (const T& x : v) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << '}';
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << '[';
  bool was = false;
  for (const T& x : v) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << ']';
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << '[';
  bool was = false;
  for (const T& x : v) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << ']';
  return os;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const map<T1, T2>& m) {
  os << '<';
  bool was = false;
  for (const auto& x : m) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << '>';
  return os;
}
const long long INF = (long long)2e9;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)2e5 + 10;
struct st {
  int a[5];
  long long sum[5];
  st() {
    for (int i = 0; i < 5; i++) {
      a[i] = 0;
      sum[i] = 0;
    }
  }
  st operator+(const st& other) const {
    st a = *this;
    for (int i = 0; i < 5; i++) {
      a.a[i] += other.a[i];
      a.sum[i] += other.sum[i];
    }
    return a;
  }
};
int n, k;
vector<int> d[MAXN];
st a[MAXN];
long long ans;
st empt;
st dfs(int x, int pr) {
  st b;
  vector<st> vec;
  for (auto to : d[x]) {
    if (to != pr) {
      vec.push_back(dfs(to, x));
      b = b + vec.back();
    }
  }
  vec.push_back(empt);
  b = b + empt;
  long long cur = 0;
  for (auto it : vec) {
    st c = b;
    for (int i = 0; i < k; i++) {
      c.a[i] -= it.a[i];
      c.sum[i] -= it.sum[i];
    }
    long long coeff;
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        if (i + j == 0)
          coeff = 0;
        else {
          if (i + j <= k) {
            coeff = 1;
          } else {
            coeff = 2;
          }
        }
        cur += (long long)c.a[i] * it.sum[j];
        cur += (long long)c.sum[i] * it.a[j];
        cur += coeff * c.a[i] * it.a[j];
      }
    }
  }
  ans += cur;
  for (int i = k - 1; i >= 1; i--) {
    swap(b.a[i], b.a[i - 1]);
    swap(b.sum[i], b.sum[i - 1]);
  }
  b.sum[0] += b.a[0];
  return b;
}
int main() {
  ios_base::sync_with_stdio(false);
  empt.a[0] = 1;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    d[x].push_back(y);
    d[y].push_back(x);
  }
  dfs(1, 0);
  cout << ans / 2;
}

### Prompt

Create a solution in cpp for 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;
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T1, class T2>
T1 chmin(T1& x, const T2& y) {
  if (T1(y) < x) return x = y;
  return x;
}
template <class T1, class T2>
T1 chmax(T1& x, const T2& y) {
  if (T1(y) > x) return x = y;
  return x;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << '{';
  bool was = false;
  for (const T& x : v) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << '}';
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << '[';
  bool was = false;
  for (const T& x : v) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << ']';
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << '[';
  bool was = false;
  for (const T& x : v) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << ']';
  return os;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const map<T1, T2>& m) {
  os << '<';
  bool was = false;
  for (const auto& x : m) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << '>';
  return os;
}
const long long INF = (long long)2e9;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)2e5 + 10;
struct st {
  int a[5];
  long long sum[5];
  st() {
    for (int i = 0; i < 5; i++) {
      a[i] = 0;
      sum[i] = 0;
    }
  }
  st operator+(const st& other) const {
    st a = *this;
    for (int i = 0; i < 5; i++) {
      a.a[i] += other.a[i];
      a.sum[i] += other.sum[i];
    }
    return a;
  }
};
int n, k;
vector<int> d[MAXN];
st a[MAXN];
long long ans;
st empt;
st dfs(int x, int pr) {
  st b;
  vector<st> vec;
  for (auto to : d[x]) {
    if (to != pr) {
      vec.push_back(dfs(to, x));
      b = b + vec.back();
    }
  }
  vec.push_back(empt);
  b = b + empt;
  long long cur = 0;
  for (auto it : vec) {
    st c = b;
    for (int i = 0; i < k; i++) {
      c.a[i] -= it.a[i];
      c.sum[i] -= it.sum[i];
    }
    long long coeff;
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        if (i + j == 0)
          coeff = 0;
        else {
          if (i + j <= k) {
            coeff = 1;
          } else {
            coeff = 2;
          }
        }
        cur += (long long)c.a[i] * it.sum[j];
        cur += (long long)c.sum[i] * it.a[j];
        cur += coeff * c.a[i] * it.a[j];
      }
    }
  }
  ans += cur;
  for (int i = k - 1; i >= 1; i--) {
    swap(b.a[i], b.a[i - 1]);
    swap(b.sum[i], b.sum[i - 1]);
  }
  b.sum[0] += b.a[0];
  return b;
}
int main() {
  ios_base::sync_with_stdio(false);
  empt.a[0] = 1;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    d[x].push_back(y);
    d[y].push_back(x);
  }
  dfs(1, 0);
  cout << ans / 2;
}

```