#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!(ch >= '0' && ch <= '9')) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * f;
}
const int N = 200005;
struct edge {
  int go, next;
} e[N << 1];
int cnt = 0, last[N];
inline void ins(int u, int v) {
  e[++cnt].go = v;
  e[cnt].next = last[u];
  last[u] = cnt;
}
int n, k;
long long dp[N][6];
int num[N][6];
long long ans = 0;
inline void dfs(int x, int fa) {
  dp[x][0] = 0;
  num[x][0] = 1;
  for (int i = last[x]; i; i = e[i].next) {
    if (e[i].go == fa) continue;
    dfs(e[i].go, x);
    for (int p = 0; p <= k - 1; p++) {
      for (int q = 0; q <= k - 1; q++) {
        ans += num[e[i].go][p] * dp[x][q] + num[x][q] * dp[e[i].go][p];
        if (p + q + 1 > k) ans += 1LL * num[e[i].go][p] * num[x][q];
        ans += 1LL * num[e[i].go][p] * num[x][q];
      }
    }
    for (int p = 0; p <= k - 1; p++) {
      dp[x][(p + 1) % k] += dp[e[i].go][p] + (p + 1 == k) * num[e[i].go][p];
      num[x][(p + 1) % k] += num[e[i].go][p];
    }
  }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n - 1; i++) {
    int u = read(), v = read();
    ins(u, v);
    ins(v, u);
  }
  dfs(1, 0);
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!(ch >= '0' && ch <= '9')) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * f;
}
const int N = 200005;
struct edge {
  int go, next;
} e[N << 1];
int cnt = 0, last[N];
inline void ins(int u, int v) {
  e[++cnt].go = v;
  e[cnt].next = last[u];
  last[u] = cnt;
}
int n, k;
long long dp[N][6];
int num[N][6];
long long ans = 0;
inline void dfs(int x, int fa) {
  dp[x][0] = 0;
  num[x][0] = 1;
  for (int i = last[x]; i; i = e[i].next) {
    if (e[i].go == fa) continue;
    dfs(e[i].go, x);
    for (int p = 0; p <= k - 1; p++) {
      for (int q = 0; q <= k - 1; q++) {
        ans += num[e[i].go][p] * dp[x][q] + num[x][q] * dp[e[i].go][p];
        if (p + q + 1 > k) ans += 1LL * num[e[i].go][p] * num[x][q];
        ans += 1LL * num[e[i].go][p] * num[x][q];
      }
    }
    for (int p = 0; p <= k - 1; p++) {
      dp[x][(p + 1) % k] += dp[e[i].go][p] + (p + 1 == k) * num[e[i].go][p];
      num[x][(p + 1) % k] += num[e[i].go][p];
    }
  }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n - 1; i++) {
    int u = read(), v = read();
    ins(u, v);
    ins(v, u);
  }
  dfs(1, 0);
  printf("%I64d\n", ans);
  return 0;
}

```