#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
const double dinf = 1e250;
const long long INF = (long long)(2e18);
const long long inf = (long long)(1e9 + 7);
const long long mod = 998244353;
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <class T>
void OUT(T a) {
  cout << a;
  exit(0);
}
template <class T1, class T2>
inline T1& maxeq(T1& a, T2 b) {
  return a = max(a, T1(b));
}
template <class T1, class T2>
inline T1& mineq(T1& a, T2 b) {
  return a = min(a, T1(b));
}
mt19937 rng((uint32_t)(chrono::steady_clock::now().time_since_epoch().count()));
chrono::time_point<chrono::steady_clock> cl;
double current_time() {
  return (chrono::steady_clock::now() - cl).count() / 1e9;
}
long long n, k;
vector<vector<long long> > g;
long long res = 0;
struct kek {
  vector<long long> nums;
  vector<long long> sums;
  kek() {}
  kek(long long i) {
    nums.resize(k, 0);
    sums.resize(k, 0);
    nums[0] = i;
  }
  kek(long long i, long long j) {
    nums.resize(k, 0);
    sums.resize(k, 0);
    nums[0] = i;
    sums[0] = j;
  }
  kek& operator+=(kek b) {
    for (long long i = 0; i < k; ++i) {
      nums[i] += b.nums[i];
      sums[i] += b.sums[i];
    }
    return *this;
  }
  kek& operator-=(kek b) {
    for (long long i = 0; i < k; ++i) {
      nums[i] -= b.nums[i];
      sums[i] -= b.sums[i];
    }
    return *this;
  }
  kek& operator*=(long long b) {
    for (auto& it : nums) it *= b;
    for (auto& it : sums) it *= b;
    return *this;
  }
  kek operator+(kek b) { return kek(*this) += b; }
  kek operator-(kek b) { return kek(*this) -= b; }
  long long operator*(kek b) {
    long long ret = 0;
    kek res(0);
    for (long long i = 0; i < k; ++i) {
      kek a1 = *this << i;
      kek a2 = b >> (i + 1);
      ret += a1.nums[0] * a2.gets();
      ret += a1.sums[0] * a2.getn();
    }
    return ret;
  }
  long long getn() {
    long long res = 0;
    for (auto& it : nums) res += it;
    return res;
  }
  long long gets() {
    long long res = 0;
    for (auto& it : sums) res += it;
    return res;
  }
  void print() {
    for (auto& it : nums) cout << it << " ";
    cout << endl;
    for (auto& it : sums) cout << it << " ";
    cout << endl;
  }
  kek operator>>(long long x) {
    kek res(0);
    for (long long i = 0; i < k; ++i) {
      res.nums[(i + x) % k] = nums[i];
      res.sums[(i + x) % k] = sums[i];
    }
    for (long long i = 1; i <= x; ++i) res.sums[i % k] += res.nums[i % k];
    return res;
  }
  kek operator<<(long long x) {
    kek res(0);
    for (long long i = 0; i < k; ++i) {
      res.nums[(i + k - x) % k] = nums[i];
      res.sums[(i + k - x) % k] = sums[i];
    }
    for (long long i = 1; i <= x; ++i)
      res.sums[(k + 1 - i) % k] -= res.nums[(k + 1 - i) % k];
    return res;
  }
};
ostream& operator<<(ostream& ostr, kek a) {
  for (auto& it : a.nums) ostr << it << " ";
  ostr << endl;
  for (auto& it : a.sums) ostr << it << " ";
  ostr << endl;
  ostr << endl;
  return ostr;
}
vector<kek> dp;
void dfs(long long v, long long p) {
  for (auto& it : g[v]) {
    if (it == p) continue;
    dfs(it, v);
    dp[v] += dp[it] >> 1;
  }
  res += dp[v].gets();
  for (auto& it : g[v]) {
    if (it == p) continue;
    kek tmp = dp[v] - (dp[it] >> 1);
    res += dp[it] * tmp;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  cl = chrono::steady_clock::now();
  cin.tie(0);
  cin >> n >> k;
  g.resize(n);
  dp.resize(n, kek(1));
  for (long long i = 0; i < n - 1; ++i) {
    long long f, t;
    cin >> f >> t;
    --f;
    --t;
    g[f].push_back(t);
    g[t].push_back(f);
  }
  dfs(0, 0);
  cout << 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 double eps = 1e-9;
const double pi = acos(-1.0);
const double dinf = 1e250;
const long long INF = (long long)(2e18);
const long long inf = (long long)(1e9 + 7);
const long long mod = 998244353;
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <class T>
void OUT(T a) {
  cout << a;
  exit(0);
}
template <class T1, class T2>
inline T1& maxeq(T1& a, T2 b) {
  return a = max(a, T1(b));
}
template <class T1, class T2>
inline T1& mineq(T1& a, T2 b) {
  return a = min(a, T1(b));
}
mt19937 rng((uint32_t)(chrono::steady_clock::now().time_since_epoch().count()));
chrono::time_point<chrono::steady_clock> cl;
double current_time() {
  return (chrono::steady_clock::now() - cl).count() / 1e9;
}
long long n, k;
vector<vector<long long> > g;
long long res = 0;
struct kek {
  vector<long long> nums;
  vector<long long> sums;
  kek() {}
  kek(long long i) {
    nums.resize(k, 0);
    sums.resize(k, 0);
    nums[0] = i;
  }
  kek(long long i, long long j) {
    nums.resize(k, 0);
    sums.resize(k, 0);
    nums[0] = i;
    sums[0] = j;
  }
  kek& operator+=(kek b) {
    for (long long i = 0; i < k; ++i) {
      nums[i] += b.nums[i];
      sums[i] += b.sums[i];
    }
    return *this;
  }
  kek& operator-=(kek b) {
    for (long long i = 0; i < k; ++i) {
      nums[i] -= b.nums[i];
      sums[i] -= b.sums[i];
    }
    return *this;
  }
  kek& operator*=(long long b) {
    for (auto& it : nums) it *= b;
    for (auto& it : sums) it *= b;
    return *this;
  }
  kek operator+(kek b) { return kek(*this) += b; }
  kek operator-(kek b) { return kek(*this) -= b; }
  long long operator*(kek b) {
    long long ret = 0;
    kek res(0);
    for (long long i = 0; i < k; ++i) {
      kek a1 = *this << i;
      kek a2 = b >> (i + 1);
      ret += a1.nums[0] * a2.gets();
      ret += a1.sums[0] * a2.getn();
    }
    return ret;
  }
  long long getn() {
    long long res = 0;
    for (auto& it : nums) res += it;
    return res;
  }
  long long gets() {
    long long res = 0;
    for (auto& it : sums) res += it;
    return res;
  }
  void print() {
    for (auto& it : nums) cout << it << " ";
    cout << endl;
    for (auto& it : sums) cout << it << " ";
    cout << endl;
  }
  kek operator>>(long long x) {
    kek res(0);
    for (long long i = 0; i < k; ++i) {
      res.nums[(i + x) % k] = nums[i];
      res.sums[(i + x) % k] = sums[i];
    }
    for (long long i = 1; i <= x; ++i) res.sums[i % k] += res.nums[i % k];
    return res;
  }
  kek operator<<(long long x) {
    kek res(0);
    for (long long i = 0; i < k; ++i) {
      res.nums[(i + k - x) % k] = nums[i];
      res.sums[(i + k - x) % k] = sums[i];
    }
    for (long long i = 1; i <= x; ++i)
      res.sums[(k + 1 - i) % k] -= res.nums[(k + 1 - i) % k];
    return res;
  }
};
ostream& operator<<(ostream& ostr, kek a) {
  for (auto& it : a.nums) ostr << it << " ";
  ostr << endl;
  for (auto& it : a.sums) ostr << it << " ";
  ostr << endl;
  ostr << endl;
  return ostr;
}
vector<kek> dp;
void dfs(long long v, long long p) {
  for (auto& it : g[v]) {
    if (it == p) continue;
    dfs(it, v);
    dp[v] += dp[it] >> 1;
  }
  res += dp[v].gets();
  for (auto& it : g[v]) {
    if (it == p) continue;
    kek tmp = dp[v] - (dp[it] >> 1);
    res += dp[it] * tmp;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  cl = chrono::steady_clock::now();
  cin.tie(0);
  cin >> n >> k;
  g.resize(n);
  dp.resize(n, kek(1));
  for (long long i = 0; i < n - 1; ++i) {
    long long f, t;
    cin >> f >> t;
    --f;
    --t;
    g[f].push_back(t);
    g[t].push_back(f);
  }
  dfs(0, 0);
  cout << res / 2;
  return 0;
}

```