#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
int n, k;
const int MAXN = 2e5 + 10;
struct node {
  int v, next;
} e[MAXN << 1];
int tot;
int head[MAXN];
void addedge(int u, int v) {
  e[tot].v = v;
  e[tot].next = head[u];
  head[u] = tot++;
}
void init() {
  tot = 0;
  memset(head, -1, sizeof head);
}
long long num[MAXN];
long long sum[MAXN];
long long ans[MAXN];
long long dp[MAXN][5];
void dfs(int u, int fa) {
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].v;
    if (v == fa) continue;
    dfs(v, u);
    ans[u] += ans[v];
    ans[u] += num[v] * sum[u] + num[u] * sum[v];
    for (int i = 0; i < k; ++i) {
      for (int j = 0; j < k; ++j) {
        int ti = i == 0 ? k : i;
        int tj = j + 1;
        int b = ti + tj;
        int jump = b / k;
        if (b % k) ++jump;
        ans[u] += dp[u][i] * dp[v][j] * jump;
      }
    }
    sum[u] += sum[v];
    num[u] += num[v];
    for (int j = 0; j < k; ++j) {
      dp[u][(j + 1) % k] += dp[v][j];
    }
  }
  sum[u] += dp[u][0];
  ans[u] += sum[u] + num[u] - dp[u][0];
  dp[u][0] += 1;
  num[u]++;
}
int main() {
  while (cin >> n >> k) {
    init();
    for (int i = 1; i < n; ++i) {
      int a, b;
      cin >> a >> b;
      addedge(a, b);
      addedge(b, a);
    }
    memset(dp, 0, sizeof dp);
    memset(sum, 0, sizeof sum);
    memset(ans, 0, sizeof ans);
    memset(num, 0, sizeof num);
    dfs(1, 1);
    cout << ans[1] << endl;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 = 0x3f3f3f3f;
int n, k;
const int MAXN = 2e5 + 10;
struct node {
  int v, next;
} e[MAXN << 1];
int tot;
int head[MAXN];
void addedge(int u, int v) {
  e[tot].v = v;
  e[tot].next = head[u];
  head[u] = tot++;
}
void init() {
  tot = 0;
  memset(head, -1, sizeof head);
}
long long num[MAXN];
long long sum[MAXN];
long long ans[MAXN];
long long dp[MAXN][5];
void dfs(int u, int fa) {
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].v;
    if (v == fa) continue;
    dfs(v, u);
    ans[u] += ans[v];
    ans[u] += num[v] * sum[u] + num[u] * sum[v];
    for (int i = 0; i < k; ++i) {
      for (int j = 0; j < k; ++j) {
        int ti = i == 0 ? k : i;
        int tj = j + 1;
        int b = ti + tj;
        int jump = b / k;
        if (b % k) ++jump;
        ans[u] += dp[u][i] * dp[v][j] * jump;
      }
    }
    sum[u] += sum[v];
    num[u] += num[v];
    for (int j = 0; j < k; ++j) {
      dp[u][(j + 1) % k] += dp[v][j];
    }
  }
  sum[u] += dp[u][0];
  ans[u] += sum[u] + num[u] - dp[u][0];
  dp[u][0] += 1;
  num[u]++;
}
int main() {
  while (cin >> n >> k) {
    init();
    for (int i = 1; i < n; ++i) {
      int a, b;
      cin >> a >> b;
      addedge(a, b);
      addedge(b, a);
    }
    memset(dp, 0, sizeof dp);
    memset(sum, 0, sizeof sum);
    memset(ans, 0, sizeof ans);
    memset(num, 0, sizeof num);
    dfs(1, 1);
    cout << ans[1] << endl;
  }
  return 0;
}

```