#include <bits/stdc++.h>
using namespace std;
const int maxn = 200000;
vector<int> g[maxn];
int64_t res[maxn];
int64_t pres[maxn];
int sz[maxn];
int k;
int n;
int rsd[maxn][5];
void dfs(int v, int p) {
  res[v] = 0;
  pres[v] = 0;
  sz[v] = 1;
  rsd[v][0] = 1;
  for (int i = 1; i < k; ++i) rsd[v][i] = 0;
  for (int u : g[v]) {
    if (u == p) continue;
    dfs(u, v);
    res[v] += res[u];
    res[v] += pres[u] * 1LL * sz[v];
    res[v] += pres[v] * 1LL * sz[u];
    res[v] += rsd[v][0] * 1LL * rsd[u][0];
    for (int i = 1; i < k; ++i)
      for (int j = 1; j < k; ++j)
        if (i + j > k) res[v] -= rsd[v][i] * 1LL * rsd[u][j];
    sz[v] += sz[u];
    pres[v] += pres[u];
    for (int i = 1; i < k; ++i) {
      rsd[v][i - 1] += rsd[u][i];
    }
    pres[v] += rsd[u][0];
    rsd[v][k - 1] += rsd[u][0];
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, -1);
  cout << res[0] << '\n';
  return 0;
}

### 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;
const int maxn = 200000;
vector<int> g[maxn];
int64_t res[maxn];
int64_t pres[maxn];
int sz[maxn];
int k;
int n;
int rsd[maxn][5];
void dfs(int v, int p) {
  res[v] = 0;
  pres[v] = 0;
  sz[v] = 1;
  rsd[v][0] = 1;
  for (int i = 1; i < k; ++i) rsd[v][i] = 0;
  for (int u : g[v]) {
    if (u == p) continue;
    dfs(u, v);
    res[v] += res[u];
    res[v] += pres[u] * 1LL * sz[v];
    res[v] += pres[v] * 1LL * sz[u];
    res[v] += rsd[v][0] * 1LL * rsd[u][0];
    for (int i = 1; i < k; ++i)
      for (int j = 1; j < k; ++j)
        if (i + j > k) res[v] -= rsd[v][i] * 1LL * rsd[u][j];
    sz[v] += sz[u];
    pres[v] += pres[u];
    for (int i = 1; i < k; ++i) {
      rsd[v][i - 1] += rsd[u][i];
    }
    pres[v] += rsd[u][0];
    rsd[v][k - 1] += rsd[u][0];
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, -1);
  cout << res[0] << '\n';
  return 0;
}

```