#include <bits/stdc++.h>
using namespace std;
int casex, n, k;
int num, hed[200010], nex[400010], vt[400010];
long long dp[200010][10], sum[200010], ans;
int read() {
  int u = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') u = (u << 1) + (u << 3) + c - '0', c = getchar();
  return u;
}
void add(int x, int y) {
  vt[++num] = y, nex[num] = hed[x], hed[x] = num;
  vt[++num] = x, nex[num] = hed[y], hed[y] = num;
}
void dfs1(int u, int fa) {
  for (int i = hed[u]; i; i = nex[i]) {
    if (vt[i] == fa) continue;
    dfs1(vt[i], u);
    for (int j = 1; j < k; j++) dp[u][j] += dp[vt[i]][j - 1];
    sum[u] += sum[vt[i]] + dp[vt[i]][k - 1];
    dp[u][0] += dp[vt[i]][k - 1];
  }
  dp[u][0]++;
}
void dfs2(int u, int fa) {
  int tmp1 = dp[u][k - 1], tmp2 = dp[u][max(0, k - 2)];
  for (int j = k - 1; j >= 2; j--) dp[u][j] += dp[fa][j - 1] - dp[u][j - 2];
  dp[u][1] += dp[fa][0] - tmp1;
  dp[u][0] += dp[fa][k - 1] - tmp2;
  sum[u] += sum[fa] - sum[u] - tmp1 + dp[fa][k - 1] - tmp2;
  for (int i = hed[u]; i; i = nex[i]) {
    if (vt[i] == fa) continue;
    dfs2(vt[i], u);
  }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n - 1; i++) {
    int x = read(), y = read();
    add(x, y);
  }
  dfs1(1, 0);
  for (int i = hed[1]; i; i = nex[i]) dfs2(vt[i], 1);
  for (int i = 1; i <= n; i++) {
    ans += sum[i];
    for (int j = 1; j < k; j++) ans += dp[i][j];
  }
  printf("%lld", ans / 2);
}

### Prompt

Generate 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;
int casex, n, k;
int num, hed[200010], nex[400010], vt[400010];
long long dp[200010][10], sum[200010], ans;
int read() {
  int u = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') u = (u << 1) + (u << 3) + c - '0', c = getchar();
  return u;
}
void add(int x, int y) {
  vt[++num] = y, nex[num] = hed[x], hed[x] = num;
  vt[++num] = x, nex[num] = hed[y], hed[y] = num;
}
void dfs1(int u, int fa) {
  for (int i = hed[u]; i; i = nex[i]) {
    if (vt[i] == fa) continue;
    dfs1(vt[i], u);
    for (int j = 1; j < k; j++) dp[u][j] += dp[vt[i]][j - 1];
    sum[u] += sum[vt[i]] + dp[vt[i]][k - 1];
    dp[u][0] += dp[vt[i]][k - 1];
  }
  dp[u][0]++;
}
void dfs2(int u, int fa) {
  int tmp1 = dp[u][k - 1], tmp2 = dp[u][max(0, k - 2)];
  for (int j = k - 1; j >= 2; j--) dp[u][j] += dp[fa][j - 1] - dp[u][j - 2];
  dp[u][1] += dp[fa][0] - tmp1;
  dp[u][0] += dp[fa][k - 1] - tmp2;
  sum[u] += sum[fa] - sum[u] - tmp1 + dp[fa][k - 1] - tmp2;
  for (int i = hed[u]; i; i = nex[i]) {
    if (vt[i] == fa) continue;
    dfs2(vt[i], u);
  }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n - 1; i++) {
    int x = read(), y = read();
    add(x, y);
  }
  dfs1(1, 0);
  for (int i = hed[1]; i; i = nex[i]) dfs2(vt[i], 1);
  for (int i = 1; i <= n; i++) {
    ans += sum[i];
    for (int j = 1; j < k; j++) ans += dp[i][j];
  }
  printf("%lld", ans / 2);
}

```