#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
bool Check(int N, int pos) { return (bool)(N & (1 << pos)); }
int Set(int N, int pos) { return (N | (1 << pos)); }
int fx[] = {-1, 0, 1, 0};
int fy[] = {0, 1, 0, -1};
inline long long BigMod(long long B, long long P, long long M) {
  long long R = 1;
  while (P > 0) {
    if (P % 2 == 1) {
      R = (R * B) % M;
    }
    P /= 2;
    B = (B * B) % M;
  }
  return R;
}
const int N = 200000;
long long n, k;
vector<long long> G[N + 5];
long long childdis[N + 5][5];
long long childdis2[N + 5][5];
long long otherdis[N + 5][5];
long long sum1[N + 5];
long long sum2[N + 5];
long long child[N + 5];
long long answer[N + 5];
long long dfs1(long long node, long long pre) {
  long long ret = 0;
  childdis[node][0] = 1;
  for (long long i = 0; i < G[node].size(); i++) {
    long long go = G[node][i];
    if (go == pre) continue;
    long long ch = dfs1(go, node);
    ret += ch;
    sum1[node] += (sum1[go] + ch);
    for (long long j = 0; j < k; j++) {
      childdis[node][(j + 1) % k] += childdis[go][j];
    }
  }
  return child[node] = ret + 1;
}
void dfs2(long long node, long long pre) {
  if (node == 1) {
    sum2[node] = sum1[node];
    for (long long i = 0; i < k; i++) childdis2[node][i] = childdis[node][i];
  }
  for (long long go : G[node]) {
    if (go == pre) continue;
    sum2[go] =
        sum1[go] + (sum2[node] - sum1[go] - (child[go])) + (n - (child[go]));
    for (long long i = 0; i < k; i++) {
      otherdis[go][i] = childdis2[node][i] - childdis[go][(i - 1 + k) % k];
    }
    for (long long i = 0; i < k; i++) {
      childdis2[go][i] = childdis[go][i];
    }
    for (long long i = 0; i < k; i++) {
      childdis2[go][(i + 1) % k] += otherdis[go][i];
    }
    dfs2(go, node);
  }
}
int main() {
  scanf("%lld %lld", &n, &k);
  for (long long i = 1; i < n; i++) {
    long long a, b;
    scanf("%lld %lld", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    long long s = sum2[i];
    for (long long j = 1; j < k; j++) {
      long long add = k - j;
      s += childdis2[i][j] * add;
    }
    ans += s / k;
  }
  printf("%lld\n", ans / 2);
}

### Prompt

Please formulate 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 long long MOD = 1e9 + 7;
bool Check(int N, int pos) { return (bool)(N & (1 << pos)); }
int Set(int N, int pos) { return (N | (1 << pos)); }
int fx[] = {-1, 0, 1, 0};
int fy[] = {0, 1, 0, -1};
inline long long BigMod(long long B, long long P, long long M) {
  long long R = 1;
  while (P > 0) {
    if (P % 2 == 1) {
      R = (R * B) % M;
    }
    P /= 2;
    B = (B * B) % M;
  }
  return R;
}
const int N = 200000;
long long n, k;
vector<long long> G[N + 5];
long long childdis[N + 5][5];
long long childdis2[N + 5][5];
long long otherdis[N + 5][5];
long long sum1[N + 5];
long long sum2[N + 5];
long long child[N + 5];
long long answer[N + 5];
long long dfs1(long long node, long long pre) {
  long long ret = 0;
  childdis[node][0] = 1;
  for (long long i = 0; i < G[node].size(); i++) {
    long long go = G[node][i];
    if (go == pre) continue;
    long long ch = dfs1(go, node);
    ret += ch;
    sum1[node] += (sum1[go] + ch);
    for (long long j = 0; j < k; j++) {
      childdis[node][(j + 1) % k] += childdis[go][j];
    }
  }
  return child[node] = ret + 1;
}
void dfs2(long long node, long long pre) {
  if (node == 1) {
    sum2[node] = sum1[node];
    for (long long i = 0; i < k; i++) childdis2[node][i] = childdis[node][i];
  }
  for (long long go : G[node]) {
    if (go == pre) continue;
    sum2[go] =
        sum1[go] + (sum2[node] - sum1[go] - (child[go])) + (n - (child[go]));
    for (long long i = 0; i < k; i++) {
      otherdis[go][i] = childdis2[node][i] - childdis[go][(i - 1 + k) % k];
    }
    for (long long i = 0; i < k; i++) {
      childdis2[go][i] = childdis[go][i];
    }
    for (long long i = 0; i < k; i++) {
      childdis2[go][(i + 1) % k] += otherdis[go][i];
    }
    dfs2(go, node);
  }
}
int main() {
  scanf("%lld %lld", &n, &k);
  for (long long i = 1; i < n; i++) {
    long long a, b;
    scanf("%lld %lld", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    long long s = sum2[i];
    for (long long j = 1; j < k; j++) {
      long long add = k - j;
      s += childdis2[i][j] * add;
    }
    ans += s / k;
  }
  printf("%lld\n", ans / 2);
}

```