#include <bits/stdc++.h>
using namespace std;
template <class T>
T sqr(const T &a) {
  return a * a;
}
const int MAXN = 2e5 + 100;
vector<int> a[MAXN];
long long res[MAXN][10], ans;
int cnt[MAXN][10];
int n, k;
void dfs(int v, int p = 0) {
  cnt[v][0] = 1;
  for (int to : a[v]) {
    if (to == p) continue;
    dfs(to, v);
    for (int j = 0; j < (k); j++)
      cnt[v][j + 1] += cnt[to][j], res[v][j + 1] += res[to][j];
  }
  cnt[v][0] += cnt[v][k];
  res[v][1] += cnt[v][1];
  res[v][0] = res[v][k];
}
void dfs1(int v, int p, vector<int> pCnt, long long pAns) {
  ans += pAns;
  for (int j = 0; j < (k); j++) ans += res[v][j];
  for (int to : a[v]) {
    if (to == p) continue;
    vector<int> newPCnt(k, 0);
    for (int i = 0; i < (k); i++) {
      newPCnt[(i + 1) % k] = pCnt[i] + cnt[v][i] - cnt[to][(k + i - 1) % k];
    }
    long long newAns = pAns;
    for (int j = 0; j < (k); j++) newAns += res[v][j] - res[to][j];
    dfs1(to, v, newPCnt,
         newAns + pCnt[0] + cnt[v][0] - cnt[to][k - 1] - cnt[to][0]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  scanf("%d%d", &n, &k);
  for (int i = 0; i < (n - 1); i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--, y--;
    a[x].push_back(y), a[y].push_back(x);
  }
  dfs(0);
  dfs1(0, 0, vector<int>(k + 1, 0), 0);
  printf("%lld\n", ans / 2);
}

### 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;
template <class T>
T sqr(const T &a) {
  return a * a;
}
const int MAXN = 2e5 + 100;
vector<int> a[MAXN];
long long res[MAXN][10], ans;
int cnt[MAXN][10];
int n, k;
void dfs(int v, int p = 0) {
  cnt[v][0] = 1;
  for (int to : a[v]) {
    if (to == p) continue;
    dfs(to, v);
    for (int j = 0; j < (k); j++)
      cnt[v][j + 1] += cnt[to][j], res[v][j + 1] += res[to][j];
  }
  cnt[v][0] += cnt[v][k];
  res[v][1] += cnt[v][1];
  res[v][0] = res[v][k];
}
void dfs1(int v, int p, vector<int> pCnt, long long pAns) {
  ans += pAns;
  for (int j = 0; j < (k); j++) ans += res[v][j];
  for (int to : a[v]) {
    if (to == p) continue;
    vector<int> newPCnt(k, 0);
    for (int i = 0; i < (k); i++) {
      newPCnt[(i + 1) % k] = pCnt[i] + cnt[v][i] - cnt[to][(k + i - 1) % k];
    }
    long long newAns = pAns;
    for (int j = 0; j < (k); j++) newAns += res[v][j] - res[to][j];
    dfs1(to, v, newPCnt,
         newAns + pCnt[0] + cnt[v][0] - cnt[to][k - 1] - cnt[to][0]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  scanf("%d%d", &n, &k);
  for (int i = 0; i < (n - 1); i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--, y--;
    a[x].push_back(y), a[y].push_back(x);
  }
  dfs(0);
  dfs1(0, 0, vector<int>(k + 1, 0), 0);
  printf("%lld\n", ans / 2);
}

```