#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2 * 1e5, MAXK = 5;
long long sum[MAXN][MAXK], ans = 0;
int N, K, cnt[MAXN][MAXK];
vector<int> G[MAXN];
void solve(int v, int p) {
  long long childsum[K];
  long long childcnt[K];
  memset(childsum, 0, sizeof(childsum));
  memset(childcnt, 0, sizeof(childcnt));
  for (int i = 0; i < (int)G[v].size(); i++) {
    int u = G[v][i];
    if (u == p) continue;
    solve(u, v);
    for (int k = 0; k < (int)K; k++) {
      childsum[k] += sum[u][k];
      childcnt[k] += cnt[u][k];
    }
  }
  for (int i = 0; i < (int)G[v].size(); i++) {
    int u = G[v][i];
    if (u == p) continue;
    for (int k = 0; k < (int)K; k++) childcnt[k] -= cnt[u][k];
    for (int k = 0; k < (int)K; k++) childsum[k] -= sum[u][k];
    for (int k1 = 0; k1 < (int)K; k1++)
      for (int k2 = 0; k2 < (int)K; k2++) {
        ans += sum[u][k1] * childcnt[k2];
        ans += cnt[u][k1] * childsum[k2];
        ans += cnt[u][k1] * childcnt[k2] * (k1 + k2 + 2 <= K ? 1 : 2);
      }
  }
  for (int i = 0; i < (int)G[v].size(); i++) {
    int u = G[v][i];
    if (u == p) continue;
    for (int k = 0; k < (int)K; k++) {
      cnt[v][(k + 1) % K] += cnt[u][k];
      sum[v][(k + 1) % K] += sum[u][k];
    }
    sum[v][0] += cnt[u][K - 1];
  }
  for (int k = 0; k < (int)K; k++) ans += sum[v][k];
  for (int k = 1; k < (int)K; k++) ans += cnt[v][k];
  cnt[v][0] += 1;
}
int main() {
  scanf("%d", &N), scanf("%d", &K);
  for (int i = 0; i < (int)N - 1; i++) {
    int v, u;
    scanf("%d", &v), scanf("%d", &u);
    v--, u--;
    G[v].push_back(u);
    G[u].push_back(v);
  }
  memset(sum, 0, sizeof(sum));
  memset(cnt, 0, sizeof(cnt));
  solve(0, -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 MAXN = 2 * 1e5, MAXK = 5;
long long sum[MAXN][MAXK], ans = 0;
int N, K, cnt[MAXN][MAXK];
vector<int> G[MAXN];
void solve(int v, int p) {
  long long childsum[K];
  long long childcnt[K];
  memset(childsum, 0, sizeof(childsum));
  memset(childcnt, 0, sizeof(childcnt));
  for (int i = 0; i < (int)G[v].size(); i++) {
    int u = G[v][i];
    if (u == p) continue;
    solve(u, v);
    for (int k = 0; k < (int)K; k++) {
      childsum[k] += sum[u][k];
      childcnt[k] += cnt[u][k];
    }
  }
  for (int i = 0; i < (int)G[v].size(); i++) {
    int u = G[v][i];
    if (u == p) continue;
    for (int k = 0; k < (int)K; k++) childcnt[k] -= cnt[u][k];
    for (int k = 0; k < (int)K; k++) childsum[k] -= sum[u][k];
    for (int k1 = 0; k1 < (int)K; k1++)
      for (int k2 = 0; k2 < (int)K; k2++) {
        ans += sum[u][k1] * childcnt[k2];
        ans += cnt[u][k1] * childsum[k2];
        ans += cnt[u][k1] * childcnt[k2] * (k1 + k2 + 2 <= K ? 1 : 2);
      }
  }
  for (int i = 0; i < (int)G[v].size(); i++) {
    int u = G[v][i];
    if (u == p) continue;
    for (int k = 0; k < (int)K; k++) {
      cnt[v][(k + 1) % K] += cnt[u][k];
      sum[v][(k + 1) % K] += sum[u][k];
    }
    sum[v][0] += cnt[u][K - 1];
  }
  for (int k = 0; k < (int)K; k++) ans += sum[v][k];
  for (int k = 1; k < (int)K; k++) ans += cnt[v][k];
  cnt[v][0] += 1;
}
int main() {
  scanf("%d", &N), scanf("%d", &K);
  for (int i = 0; i < (int)N - 1; i++) {
    int v, u;
    scanf("%d", &v), scanf("%d", &u);
    v--, u--;
    G[v].push_back(u);
    G[u].push_back(v);
  }
  memset(sum, 0, sizeof(sum));
  memset(cnt, 0, sizeof(cnt));
  solve(0, -1);
  printf("%lld\n", ans);
}

```