#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
struct edge {
  int v, nxt;
} e[N << 1];
int n, k, ect, head[N];
long long siz[N], dp[N][5], ans;
void adde(int u, int v) {
  e[ect].v = v;
  e[ect].nxt = head[u];
  head[u] = ect++;
}
void dfs(int cur, int pre) {
  siz[cur]++, dp[cur][0]++;
  for (int i = head[cur]; ~i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == pre) continue;
    dfs(v, cur);
    siz[cur] += siz[v];
    ans += siz[v] * (n - siz[v]);
    for (int j = 0; j < k; j++)
      for (int h = 0; h < k; h++) {
        int m = (h + j + 1) % k;
        if (m) m = k - m;
        ans += dp[cur][j] * dp[v][h] * m;
      }
    for (int j = 0; j < k; j++) dp[cur][j] += dp[v][(j - 1 + k) % k];
  }
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adde(u, v), adde(v, u);
  }
  dfs(1, 0);
  printf("%lld", ans / k);
}

### 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;
const int N = 2e5 + 5;
struct edge {
  int v, nxt;
} e[N << 1];
int n, k, ect, head[N];
long long siz[N], dp[N][5], ans;
void adde(int u, int v) {
  e[ect].v = v;
  e[ect].nxt = head[u];
  head[u] = ect++;
}
void dfs(int cur, int pre) {
  siz[cur]++, dp[cur][0]++;
  for (int i = head[cur]; ~i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == pre) continue;
    dfs(v, cur);
    siz[cur] += siz[v];
    ans += siz[v] * (n - siz[v]);
    for (int j = 0; j < k; j++)
      for (int h = 0; h < k; h++) {
        int m = (h + j + 1) % k;
        if (m) m = k - m;
        ans += dp[cur][j] * dp[v][h] * m;
      }
    for (int j = 0; j < k; j++) dp[cur][j] += dp[v][(j - 1 + k) % k];
  }
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adde(u, v), adde(v, u);
  }
  dfs(1, 0);
  printf("%lld", ans / k);
}

```