#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
bool used[maxn];
int n, k;
struct Node {
  long long cnt;
  long long mod[5];
  Node() : cnt(0) {
    for (int i = 0; i < k; i++) mod[i] = 0;
  }
} nodes[maxn];
vector<int> graph[maxn];
void dfs_up(int u) {
  used[u] = true;
  nodes[u].mod[0] = 1;
  for (int v : graph[u]) {
    if (!used[v]) {
      dfs_up(v);
      nodes[u].cnt += nodes[v].cnt + nodes[v].mod[k - 1];
      for (int i = 0; i < k; i++) {
        nodes[u].mod[(i + 1) % k] += nodes[v].mod[i];
      }
    }
  }
}
void dfs_down(int u) {
  used[u] = true;
  for (int v : graph[u]) {
    if (!used[v]) {
      long long tmp[5];
      long long tmpcnt = nodes[u].cnt;
      for (int i = 0; i < k; i++) tmp[i] = nodes[u].mod[i];
      tmpcnt -= nodes[v].cnt + nodes[v].mod[k - 1];
      for (int i = 0; i < k; i++) {
        tmp[(i + 1) % k] -= nodes[v].mod[i];
      }
      nodes[v].cnt += tmpcnt + tmp[k - 1];
      for (int i = 0; i < k; i++) {
        nodes[v].mod[(i + 1) % k] += tmp[i];
      }
      dfs_down(v);
    }
  }
}
long long dfs_cnt(int u) {
  used[u] = true;
  long long ans = nodes[u].cnt;
  for (int i = 1; i < k; i++) {
    ans += nodes[u].mod[i];
  }
  for (int v : graph[u]) {
    if (!used[v]) {
      ans += dfs_cnt(v);
    }
  }
  return ans;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  memset(used, 0, sizeof(used));
  dfs_up(0);
  memset(used, 0, sizeof(used));
  dfs_down(0);
  memset(used, 0, sizeof(used));
  cout << dfs_cnt(0) / 2;
  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 maxn = 200005;
bool used[maxn];
int n, k;
struct Node {
  long long cnt;
  long long mod[5];
  Node() : cnt(0) {
    for (int i = 0; i < k; i++) mod[i] = 0;
  }
} nodes[maxn];
vector<int> graph[maxn];
void dfs_up(int u) {
  used[u] = true;
  nodes[u].mod[0] = 1;
  for (int v : graph[u]) {
    if (!used[v]) {
      dfs_up(v);
      nodes[u].cnt += nodes[v].cnt + nodes[v].mod[k - 1];
      for (int i = 0; i < k; i++) {
        nodes[u].mod[(i + 1) % k] += nodes[v].mod[i];
      }
    }
  }
}
void dfs_down(int u) {
  used[u] = true;
  for (int v : graph[u]) {
    if (!used[v]) {
      long long tmp[5];
      long long tmpcnt = nodes[u].cnt;
      for (int i = 0; i < k; i++) tmp[i] = nodes[u].mod[i];
      tmpcnt -= nodes[v].cnt + nodes[v].mod[k - 1];
      for (int i = 0; i < k; i++) {
        tmp[(i + 1) % k] -= nodes[v].mod[i];
      }
      nodes[v].cnt += tmpcnt + tmp[k - 1];
      for (int i = 0; i < k; i++) {
        nodes[v].mod[(i + 1) % k] += tmp[i];
      }
      dfs_down(v);
    }
  }
}
long long dfs_cnt(int u) {
  used[u] = true;
  long long ans = nodes[u].cnt;
  for (int i = 1; i < k; i++) {
    ans += nodes[u].mod[i];
  }
  for (int v : graph[u]) {
    if (!used[v]) {
      ans += dfs_cnt(v);
    }
  }
  return ans;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  memset(used, 0, sizeof(used));
  dfs_up(0);
  memset(used, 0, sizeof(used));
  dfs_down(0);
  memset(used, 0, sizeof(used));
  cout << dfs_cnt(0) / 2;
  return 0;
}

```