#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3c3c3c3c;
const long long LINF = 0x3c3c3c3c3c3c3c3cL;
const int maxN = 1 << 18;
const int maxK = 5;
vector<int> g[maxN];
int k;
struct DP {
  long long sum[maxK];
  long long cnt[maxK];
  DP() {
    memset(sum, 0, sizeof(sum));
    memset(cnt, 0, sizeof(cnt));
  }
  DP up() {
    DP res;
    for (int o = 0; o < k; o++) {
      int o2 = (o + 1) % k;
      res.cnt[o2] += cnt[o];
      res.sum[o2] += sum[o];
      if (o + 1 >= k) {
        res.sum[o2] += cnt[o];
      }
    }
    return res;
  }
  DP& operator+=(DP& other) {
    for (int o = 0; o < k; o++) {
      cnt[o] += other.cnt[o];
      sum[o] += other.sum[o];
    }
    return *this;
  }
  DP& operator-=(DP& other) {
    for (int o = 0; o < k; o++) {
      cnt[o] -= other.cnt[o];
      sum[o] -= other.sum[o];
    }
    return *this;
  }
};
DP dp[maxN];
long long ans = 0;
void dfs(int v, int p) {
  for (auto u : g[v]) {
    if (u == p) {
      continue;
    }
    dfs(u, v);
    DP cur = dp[u].up();
    dp[v] += cur;
  }
  for (auto u : g[v]) {
    if (u == p) {
      continue;
    }
    DP dp_v = dp[v];
    DP cur = dp[u].up();
    dp_v -= cur;
    for (int o_cur = 0; o_cur < k; o_cur++) {
      for (int o_v = 0; o_v < k; o_v++) {
        ans += dp_v.sum[o_v] * cur.cnt[o_cur] + cur.sum[o_cur] * dp_v.cnt[o_v];
        ans +=
            max(0, (o_v + o_cur + k - 1) / k) * cur.cnt[o_cur] * dp_v.cnt[o_v];
      }
    }
  }
  for (int o = 0; o < k; o++) {
    ans += 2 * (dp[v].sum[o] + dp[v].cnt[o] * (int(o > 0)));
  }
  dp[v].cnt[0]++;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(1, 0);
  cout << ans / 2 << endl;
  return 0;
}

### 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;
const int INF = 0x3c3c3c3c;
const long long LINF = 0x3c3c3c3c3c3c3c3cL;
const int maxN = 1 << 18;
const int maxK = 5;
vector<int> g[maxN];
int k;
struct DP {
  long long sum[maxK];
  long long cnt[maxK];
  DP() {
    memset(sum, 0, sizeof(sum));
    memset(cnt, 0, sizeof(cnt));
  }
  DP up() {
    DP res;
    for (int o = 0; o < k; o++) {
      int o2 = (o + 1) % k;
      res.cnt[o2] += cnt[o];
      res.sum[o2] += sum[o];
      if (o + 1 >= k) {
        res.sum[o2] += cnt[o];
      }
    }
    return res;
  }
  DP& operator+=(DP& other) {
    for (int o = 0; o < k; o++) {
      cnt[o] += other.cnt[o];
      sum[o] += other.sum[o];
    }
    return *this;
  }
  DP& operator-=(DP& other) {
    for (int o = 0; o < k; o++) {
      cnt[o] -= other.cnt[o];
      sum[o] -= other.sum[o];
    }
    return *this;
  }
};
DP dp[maxN];
long long ans = 0;
void dfs(int v, int p) {
  for (auto u : g[v]) {
    if (u == p) {
      continue;
    }
    dfs(u, v);
    DP cur = dp[u].up();
    dp[v] += cur;
  }
  for (auto u : g[v]) {
    if (u == p) {
      continue;
    }
    DP dp_v = dp[v];
    DP cur = dp[u].up();
    dp_v -= cur;
    for (int o_cur = 0; o_cur < k; o_cur++) {
      for (int o_v = 0; o_v < k; o_v++) {
        ans += dp_v.sum[o_v] * cur.cnt[o_cur] + cur.sum[o_cur] * dp_v.cnt[o_v];
        ans +=
            max(0, (o_v + o_cur + k - 1) / k) * cur.cnt[o_cur] * dp_v.cnt[o_v];
      }
    }
  }
  for (int o = 0; o < k; o++) {
    ans += 2 * (dp[v].sum[o] + dp[v].cnt[o] * (int(o > 0)));
  }
  dp[v].cnt[0]++;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(1, 0);
  cout << ans / 2 << endl;
  return 0;
}

```