#include <bits/stdc++.h>
using namespace std;
const int N = 200100;
const int M = 1010;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b /= 2;
  }
  return ans;
}
long long qpow(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % c;
    a = a * a % c;
    b /= 2;
  }
  return ans;
}
int n, k, head[N], tot;
long long num[N][5], step[N][5];
long long ans = 0;
struct EDGE {
  int to, nex;
} e[N * 2];
void add(int u, int v) {
  e[tot].to = v;
  e[tot].nex = head[u];
  head[u] = tot++;
}
void dfs(int x, int fa) {
  num[x][0]++;
  for (int i = head[x]; i != -1; i = e[i].nex) {
    if (e[i].to == fa) continue;
    int y = e[i].to;
    dfs(y, x);
    for (int j = 0; j < k; j++)
      for (int l = 0; l < k; l++) {
        int s = (j + l + 1) % k == 0 ? (j + l + 1) / k : (j + l + 1) / k + 1;
        ans += 1ll * num[x][j] * step[y][l] + 1ll * num[y][l] * step[x][j] +
               1ll * s * num[x][j] * num[y][l];
      }
    for (int j = 0; j < k; j++) num[x][j] += num[y][(j + k - 1) % k];
    for (int j = 1; j < k; j++) step[x][j] += step[y][j - 1];
    step[x][0] += step[y][k - 1] + num[y][k - 1];
  }
}
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);
    add(u, v);
    add(v, u);
  }
  dfs(1, 1);
  printf("%lld\n", ans);
}

### Prompt

In Cpp, your task is to solve 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 = 200100;
const int M = 1010;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b /= 2;
  }
  return ans;
}
long long qpow(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % c;
    a = a * a % c;
    b /= 2;
  }
  return ans;
}
int n, k, head[N], tot;
long long num[N][5], step[N][5];
long long ans = 0;
struct EDGE {
  int to, nex;
} e[N * 2];
void add(int u, int v) {
  e[tot].to = v;
  e[tot].nex = head[u];
  head[u] = tot++;
}
void dfs(int x, int fa) {
  num[x][0]++;
  for (int i = head[x]; i != -1; i = e[i].nex) {
    if (e[i].to == fa) continue;
    int y = e[i].to;
    dfs(y, x);
    for (int j = 0; j < k; j++)
      for (int l = 0; l < k; l++) {
        int s = (j + l + 1) % k == 0 ? (j + l + 1) / k : (j + l + 1) / k + 1;
        ans += 1ll * num[x][j] * step[y][l] + 1ll * num[y][l] * step[x][j] +
               1ll * s * num[x][j] * num[y][l];
      }
    for (int j = 0; j < k; j++) num[x][j] += num[y][(j + k - 1) % k];
    for (int j = 1; j < k; j++) step[x][j] += step[y][j - 1];
    step[x][0] += step[y][k - 1] + num[y][k - 1];
  }
}
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);
    add(u, v);
    add(v, u);
  }
  dfs(1, 1);
  printf("%lld\n", ans);
}

```