#include <bits/stdc++.h>
using namespace std;
const int INF = 2 * int(1e9);
const long long INFll = 1ll * int(1e9) * int(1e9);
const int MOD = 1000000007;
template <typename T, typename Q>
ostream& operator<<(ostream& out, pair<T, Q>& a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename Q>
istream& operator>>(istream& in, pair<T, Q>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) in >> a[i];
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (int i = 0; i < a.size(); ++i)
    if (i == a.size() - 1)
      out << a[i];
    else
      out << a[i] << " ";
  return out;
}
const int N = 200005;
int n, k;
long long dp[N][6];
long long up[N][6];
int cnt[N][6];
int upcnt[N][6];
vector<int> g[N];
long long ans = 0;
void dfs(int v, int anc = -1) {
  cnt[v][0] = 1;
  for (int i = 0; i < k; ++i) dp[v][i] = 0;
  for (int to : g[v])
    if (to != anc) {
      dfs(to, v);
      for (int i = 0; i < k; ++i) cnt[v][(i + 1) % k] += cnt[to][i];
      for (int i = 0; i < k; ++i)
        dp[v][(i + 1) % k] += dp[to][i] + cnt[to][i] * (i == 0);
    }
}
void dfs2(int v, int anc = -1) {
  for (int i = 0; i < k; ++i) ans += dp[v][i] + up[v][i];
  upcnt[v][0] += 1;
  for (int to : g[v])
    if (to != anc)
      for (int i = 0; i < k; ++i) {
        up[v][(i + 1) % k] += dp[to][i] + cnt[to][i] * (i == 0);
        upcnt[v][(i + 1) % k] += cnt[to][i];
      }
  for (int to : g[v]) {
    if (to != anc) {
      for (int i = 0; i < k; ++i) {
        up[v][(i + 1) % k] -= dp[to][i] + cnt[to][i] * (i == 0);
        upcnt[v][(i + 1) % k] -= cnt[to][i];
      }
      for (int i = 0; i < k; ++i) {
        up[to][(i + 1) % k] += up[v][i] + upcnt[v][i] * (i == 0);
        upcnt[to][(i + 1) % k] += upcnt[v][i];
      }
      dfs2(to, v);
      for (int i = 0; i < k; ++i) {
        up[v][(i + 1) % k] += dp[to][i] + cnt[to][i] * (i == 0);
        upcnt[v][(i + 1) % k] += cnt[to][i];
      }
    }
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  dfs2(0);
  if (0) cerr << ans << endl;
  cout << ans / 2;
  if (0)
    cerr << fixed << setprecision(0)
         << "TIME = " << clock() / (long double)CLOCKS_PER_SEC * 1000 << "\n";
  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 INF = 2 * int(1e9);
const long long INFll = 1ll * int(1e9) * int(1e9);
const int MOD = 1000000007;
template <typename T, typename Q>
ostream& operator<<(ostream& out, pair<T, Q>& a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename Q>
istream& operator>>(istream& in, pair<T, Q>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) in >> a[i];
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (int i = 0; i < a.size(); ++i)
    if (i == a.size() - 1)
      out << a[i];
    else
      out << a[i] << " ";
  return out;
}
const int N = 200005;
int n, k;
long long dp[N][6];
long long up[N][6];
int cnt[N][6];
int upcnt[N][6];
vector<int> g[N];
long long ans = 0;
void dfs(int v, int anc = -1) {
  cnt[v][0] = 1;
  for (int i = 0; i < k; ++i) dp[v][i] = 0;
  for (int to : g[v])
    if (to != anc) {
      dfs(to, v);
      for (int i = 0; i < k; ++i) cnt[v][(i + 1) % k] += cnt[to][i];
      for (int i = 0; i < k; ++i)
        dp[v][(i + 1) % k] += dp[to][i] + cnt[to][i] * (i == 0);
    }
}
void dfs2(int v, int anc = -1) {
  for (int i = 0; i < k; ++i) ans += dp[v][i] + up[v][i];
  upcnt[v][0] += 1;
  for (int to : g[v])
    if (to != anc)
      for (int i = 0; i < k; ++i) {
        up[v][(i + 1) % k] += dp[to][i] + cnt[to][i] * (i == 0);
        upcnt[v][(i + 1) % k] += cnt[to][i];
      }
  for (int to : g[v]) {
    if (to != anc) {
      for (int i = 0; i < k; ++i) {
        up[v][(i + 1) % k] -= dp[to][i] + cnt[to][i] * (i == 0);
        upcnt[v][(i + 1) % k] -= cnt[to][i];
      }
      for (int i = 0; i < k; ++i) {
        up[to][(i + 1) % k] += up[v][i] + upcnt[v][i] * (i == 0);
        upcnt[to][(i + 1) % k] += upcnt[v][i];
      }
      dfs2(to, v);
      for (int i = 0; i < k; ++i) {
        up[v][(i + 1) % k] += dp[to][i] + cnt[to][i] * (i == 0);
        upcnt[v][(i + 1) % k] += cnt[to][i];
      }
    }
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  dfs2(0);
  if (0) cerr << ans << endl;
  cout << ans / 2;
  if (0)
    cerr << fixed << setprecision(0)
         << "TIME = " << clock() / (long double)CLOCKS_PER_SEC * 1000 << "\n";
  return 0;
}

```