#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200010;
const int MAXK = 6;
const int INF = 0x3f3f3f3f;
class LFS {
 public:
  LFS() {
    memset(head, -1, sizeof head);
    ecnt = 0;
    n = 0;
  }
  LFS(int N) {
    memset(head, -1, sizeof head);
    ecnt = 0;
    n = N;
  }
  void adde(int from, int to, int w) {
    e[ecnt].to = to;
    e[ecnt].w = w;
    e[ecnt].next = head[from];
    head[from] = ecnt++;
  }
  void addde(int a, int b, int w) {
    adde(a, b, w);
    adde(b, a, w);
  }
  long long solve(int k) {
    this->k = k;
    ans = 0;
    memset(dp, 0, sizeof dp);
    memset(sz, 0, sizeof sz);
    dfs(1, 0);
    return ans;
  }

 protected:
  struct Edge {
    int to, next, w;
  } e[MAXN * 2];
  int head[MAXN];
  int ecnt;
  int n, k;
  long long ans;
  long long dp[MAXN][MAXK];
  long long sz[MAXN][MAXK];

 private:
  void dfs(int u, int fa) {
    for (int i = head[u]; i != -1; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) {
        dfs(v, u);
        long long dpv[MAXK] = {0};
        long long szv[MAXK] = {0};
        for (int j = 0; j < k; j++) {
          szv[(j + 1) % k] += sz[v][j];
        }
        for (int j = 1; j < k; j++) {
          dpv[(j + 1) % k] += dp[v][j];
        }
        dpv[1 % k] += sz[v][0] + dp[v][0];
        for (int ku = 0; ku <= k; ku++) {
          for (int kv = 0; kv <= k; kv++) {
            ans += dp[u][ku] * szv[kv] + sz[u][ku] * dpv[kv];
            if (ku > 0 && kv > 0 && ku + kv <= k) {
              ans -= sz[u][ku] * szv[kv];
            }
          }
        }
        for (int j = 0; j < k; j++) {
          dp[u][j] += dpv[j];
          sz[u][j] += szv[j];
        }
      }
    }
    for (int i = 0; i < k; i++) {
      ans += dp[u][i];
    }
    sz[u][0]++;
  }
};
inline int read() {
  int X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    X = (X << 3) + (X << 1) + (ch ^ 48);
    ch = getchar();
  }
  return w ? -X : X;
}
int main() {
  int n = read();
  int k = read();
  LFS* tree = new LFS(n);
  for (int i = 1; i < n; i++) {
    int u = read();
    int v = read();
    tree->addde(u, v, 1);
  }
  cout << tree->solve(k) << endl;
}

### Prompt

Generate 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 MAXN = 200010;
const int MAXK = 6;
const int INF = 0x3f3f3f3f;
class LFS {
 public:
  LFS() {
    memset(head, -1, sizeof head);
    ecnt = 0;
    n = 0;
  }
  LFS(int N) {
    memset(head, -1, sizeof head);
    ecnt = 0;
    n = N;
  }
  void adde(int from, int to, int w) {
    e[ecnt].to = to;
    e[ecnt].w = w;
    e[ecnt].next = head[from];
    head[from] = ecnt++;
  }
  void addde(int a, int b, int w) {
    adde(a, b, w);
    adde(b, a, w);
  }
  long long solve(int k) {
    this->k = k;
    ans = 0;
    memset(dp, 0, sizeof dp);
    memset(sz, 0, sizeof sz);
    dfs(1, 0);
    return ans;
  }

 protected:
  struct Edge {
    int to, next, w;
  } e[MAXN * 2];
  int head[MAXN];
  int ecnt;
  int n, k;
  long long ans;
  long long dp[MAXN][MAXK];
  long long sz[MAXN][MAXK];

 private:
  void dfs(int u, int fa) {
    for (int i = head[u]; i != -1; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) {
        dfs(v, u);
        long long dpv[MAXK] = {0};
        long long szv[MAXK] = {0};
        for (int j = 0; j < k; j++) {
          szv[(j + 1) % k] += sz[v][j];
        }
        for (int j = 1; j < k; j++) {
          dpv[(j + 1) % k] += dp[v][j];
        }
        dpv[1 % k] += sz[v][0] + dp[v][0];
        for (int ku = 0; ku <= k; ku++) {
          for (int kv = 0; kv <= k; kv++) {
            ans += dp[u][ku] * szv[kv] + sz[u][ku] * dpv[kv];
            if (ku > 0 && kv > 0 && ku + kv <= k) {
              ans -= sz[u][ku] * szv[kv];
            }
          }
        }
        for (int j = 0; j < k; j++) {
          dp[u][j] += dpv[j];
          sz[u][j] += szv[j];
        }
      }
    }
    for (int i = 0; i < k; i++) {
      ans += dp[u][i];
    }
    sz[u][0]++;
  }
};
inline int read() {
  int X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    X = (X << 3) + (X << 1) + (ch ^ 48);
    ch = getchar();
  }
  return w ? -X : X;
}
int main() {
  int n = read();
  int k = read();
  LFS* tree = new LFS(n);
  for (int i = 1; i < n; i++) {
    int u = read();
    int v = read();
    tree->addde(u, v, 1);
  }
  cout << tree->solve(k) << endl;
}

```