#include <bits/stdc++.h>
using namespace std;
long long dp[200005][5][2];
int size[200005][5], K, size2[200005][5];
vector<int> g[200005];
void dfs(int u, int fa) {
  dp[u][0][0] = 0;
  size[u][0] = 1;
  for (int i = 0; i < (int)g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    dfs(v, u);
    for (int j = 0; j < K; j++) {
      if (j == 0) {
        dp[u][(j + 1) % K][0] += dp[v][j][0] + size[v][j];
        size[u][(j + 1) % K] += size[v][j];
      } else {
        dp[u][(j + 1) % K][0] += dp[v][j][0];
        size[u][(j + 1) % K] += size[v][j];
      }
    }
  }
}
void dfs2(int u, int fa) {
  for (int i = 0; i < (int)g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    for (int j = 0; j < K; j++) {
      if (u == 1) {
        if ((j - 1 + K) % K == 0) {
          dp[v][(j + 1) % K][1] = dp[u][j][0] - dp[v][(j - 1 + K) % K][0] -
                                  size[v][(j - 1 + K) % K];
        } else {
          dp[v][(j + 1) % K][1] = dp[u][j][0] - dp[v][(j - 1 + K) % K][0];
        }
        if (j != 0)
          size2[v][(j + 1) % K] = size[u][j] - size[v][(j - 1 + K) % K];
        else
          size2[v][(j + 1) % K] = size[u][j] - size[v][(j - 1 + K) % K] - 1;
        if (j == 0) dp[v][(j + 1) % K][1] += size2[v][(j + 1) % K];
      } else {
        dp[v][(j + 1) % K][1] = dp[u][j][1];
        if ((j - 1 + K) % K == 0) {
          dp[v][(j + 1) % K][1] += dp[u][j][0] - dp[v][(j - 1 + K) % K][0] -
                                   size[v][(j - 1 + K) % K];
        } else {
          dp[v][(j + 1) % K][1] += dp[u][j][0] - dp[v][(j - 1 + K) % K][0];
        }
        if (j != 0)
          size2[v][(j + 1) % K] =
              size2[u][j] + size[u][j] - size[v][(j - 1 + K) % K];
        else
          size2[v][(j + 1) % K] =
              size2[u][j] + size[u][j] - size[v][(j - 1 + K) % K] - 1;
        if (j == 0) dp[v][(j + 1) % K][1] += size2[v][(j + 1) % K];
      }
    }
    dfs2(v, u);
  }
}
int main() {
  int N, u, v;
  scanf("%d%d", &N, &K);
  for (int i = 1; i < N; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, -1);
  dfs2(1, -1);
  long long ans = 0, ans2 = 0;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j < K; j++) {
      ans += dp[i][j][0];
      ans2 += dp[i][j][1];
    }
  printf("%lld\n", ans + ans2 / 2);
  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;
long long dp[200005][5][2];
int size[200005][5], K, size2[200005][5];
vector<int> g[200005];
void dfs(int u, int fa) {
  dp[u][0][0] = 0;
  size[u][0] = 1;
  for (int i = 0; i < (int)g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    dfs(v, u);
    for (int j = 0; j < K; j++) {
      if (j == 0) {
        dp[u][(j + 1) % K][0] += dp[v][j][0] + size[v][j];
        size[u][(j + 1) % K] += size[v][j];
      } else {
        dp[u][(j + 1) % K][0] += dp[v][j][0];
        size[u][(j + 1) % K] += size[v][j];
      }
    }
  }
}
void dfs2(int u, int fa) {
  for (int i = 0; i < (int)g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    for (int j = 0; j < K; j++) {
      if (u == 1) {
        if ((j - 1 + K) % K == 0) {
          dp[v][(j + 1) % K][1] = dp[u][j][0] - dp[v][(j - 1 + K) % K][0] -
                                  size[v][(j - 1 + K) % K];
        } else {
          dp[v][(j + 1) % K][1] = dp[u][j][0] - dp[v][(j - 1 + K) % K][0];
        }
        if (j != 0)
          size2[v][(j + 1) % K] = size[u][j] - size[v][(j - 1 + K) % K];
        else
          size2[v][(j + 1) % K] = size[u][j] - size[v][(j - 1 + K) % K] - 1;
        if (j == 0) dp[v][(j + 1) % K][1] += size2[v][(j + 1) % K];
      } else {
        dp[v][(j + 1) % K][1] = dp[u][j][1];
        if ((j - 1 + K) % K == 0) {
          dp[v][(j + 1) % K][1] += dp[u][j][0] - dp[v][(j - 1 + K) % K][0] -
                                   size[v][(j - 1 + K) % K];
        } else {
          dp[v][(j + 1) % K][1] += dp[u][j][0] - dp[v][(j - 1 + K) % K][0];
        }
        if (j != 0)
          size2[v][(j + 1) % K] =
              size2[u][j] + size[u][j] - size[v][(j - 1 + K) % K];
        else
          size2[v][(j + 1) % K] =
              size2[u][j] + size[u][j] - size[v][(j - 1 + K) % K] - 1;
        if (j == 0) dp[v][(j + 1) % K][1] += size2[v][(j + 1) % K];
      }
    }
    dfs2(v, u);
  }
}
int main() {
  int N, u, v;
  scanf("%d%d", &N, &K);
  for (int i = 1; i < N; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, -1);
  dfs2(1, -1);
  long long ans = 0, ans2 = 0;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j < K; j++) {
      ans += dp[i][j][0];
      ans2 += dp[i][j][1];
    }
  printf("%lld\n", ans + ans2 / 2);
  return 0;
}

```