#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
int n, k;
struct edge {
  long long ans[6], mo[6];
  int to;
  int from;
  int next;
  edge() {
    next = -1;
    to = from = 0;
    memset(ans, 0, sizeof(ans));
    memset(mo, 0, sizeof(mo));
  }
} E[MAXN * 2];
struct point {
  long long ans[6];
  long long mo[6];
  point() {
    memset(ans, 0, sizeof(ans));
    memset(mo, 0, sizeof(mo));
  }
};
point A[MAXN];
int Q[MAXN * 2];
int inof[MAXN];
int deep;
void add(int v, int u) {
  E[deep].to = u;
  E[deep].from = v;
  E[deep].next = inof[v];
  inof[v] = deep++;
  E[deep].to = v;
  E[deep].from = u;
  E[deep].next = inof[u];
  inof[u] = deep++;
}
void BFS1() {
  int l = 0, r = 0;
  for (int i = inof[1]; i > -1; i = E[i].next) Q[r++] = i;
  while (l < r) {
    int id = Q[l++];
    edge &e = E[id];
    for (int i = inof[e.to]; i > -1; i = E[i].next) {
      edge &a = E[i];
      if (a.to != e.from) Q[r++] = i;
    }
  }
  r--;
  while (r > -1) {
    int id = Q[r--];
    edge &e = E[id];
    e.mo[k] = 1;
    for (int i = inof[e.to]; i > -1; i = E[i].next) {
      edge &a = E[i];
      if (a.to != e.from) {
        e.ans[1] += a.ans[k] + a.mo[k];
        e.mo[1] += a.mo[k];
        for (int t = 1; t < k; t++) {
          e.ans[t + 1] += a.ans[t];
          e.mo[t + 1] += a.mo[t];
        }
      }
    }
  }
}
void BFS2() {
  int l = 0, r = 0;
  A[1].mo[k] = 1;
  for (int i = inof[1]; i > -1; i = E[i].next) {
    A[1].ans[1] += E[i].ans[k] + E[i].mo[k];
    A[1].mo[1] += E[i].mo[k];
    for (int t = 1; t < k; t++) {
      A[1].ans[t + 1] += E[i].ans[t];
      A[1].mo[t + 1] += E[i].mo[t];
    }
    Q[r++] = i;
  }
  point bu;
  while (l < r) {
    int id = Q[l++];
    edge &e = E[id];
    A[e.to].mo[k] = 1;
    bu.ans[1] = A[e.from].ans[1] - e.mo[k] - e.ans[k];
    bu.mo[1] = A[e.from].mo[1] - e.mo[k];
    for (int t = 1; t < k; t++) {
      bu.ans[t + 1] = A[e.from].ans[t + 1] - e.ans[t];
      bu.mo[t + 1] = A[e.from].mo[t + 1] - e.mo[t];
    }
    A[e.to].ans[1] = e.ans[1] + bu.ans[k] + bu.mo[k];
    A[e.to].mo[1] = e.mo[1] + bu.mo[k];
    for (int t = 1; t < k; t++) {
      A[e.to].ans[t + 1] = e.ans[t + 1] + bu.ans[t];
      A[e.to].mo[t + 1] = e.mo[t + 1] + bu.mo[t];
    }
    for (int i = inof[e.to]; i > -1; i = E[i].next) {
      if (E[i].to != e.from) Q[r++] = i;
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  memset(inof, -1, sizeof(inof));
  long long ans = 0;
  int a, b;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    add(a, b);
  }
  BFS1();
  BFS2();
  for (int i = 1; i <= n; i++)
    for (int t = 1; t <= k; t++) ans += A[i].ans[t];
  cout << ans / 2 << endl;
  return 0;
}

### 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 = 200005;
int n, k;
struct edge {
  long long ans[6], mo[6];
  int to;
  int from;
  int next;
  edge() {
    next = -1;
    to = from = 0;
    memset(ans, 0, sizeof(ans));
    memset(mo, 0, sizeof(mo));
  }
} E[MAXN * 2];
struct point {
  long long ans[6];
  long long mo[6];
  point() {
    memset(ans, 0, sizeof(ans));
    memset(mo, 0, sizeof(mo));
  }
};
point A[MAXN];
int Q[MAXN * 2];
int inof[MAXN];
int deep;
void add(int v, int u) {
  E[deep].to = u;
  E[deep].from = v;
  E[deep].next = inof[v];
  inof[v] = deep++;
  E[deep].to = v;
  E[deep].from = u;
  E[deep].next = inof[u];
  inof[u] = deep++;
}
void BFS1() {
  int l = 0, r = 0;
  for (int i = inof[1]; i > -1; i = E[i].next) Q[r++] = i;
  while (l < r) {
    int id = Q[l++];
    edge &e = E[id];
    for (int i = inof[e.to]; i > -1; i = E[i].next) {
      edge &a = E[i];
      if (a.to != e.from) Q[r++] = i;
    }
  }
  r--;
  while (r > -1) {
    int id = Q[r--];
    edge &e = E[id];
    e.mo[k] = 1;
    for (int i = inof[e.to]; i > -1; i = E[i].next) {
      edge &a = E[i];
      if (a.to != e.from) {
        e.ans[1] += a.ans[k] + a.mo[k];
        e.mo[1] += a.mo[k];
        for (int t = 1; t < k; t++) {
          e.ans[t + 1] += a.ans[t];
          e.mo[t + 1] += a.mo[t];
        }
      }
    }
  }
}
void BFS2() {
  int l = 0, r = 0;
  A[1].mo[k] = 1;
  for (int i = inof[1]; i > -1; i = E[i].next) {
    A[1].ans[1] += E[i].ans[k] + E[i].mo[k];
    A[1].mo[1] += E[i].mo[k];
    for (int t = 1; t < k; t++) {
      A[1].ans[t + 1] += E[i].ans[t];
      A[1].mo[t + 1] += E[i].mo[t];
    }
    Q[r++] = i;
  }
  point bu;
  while (l < r) {
    int id = Q[l++];
    edge &e = E[id];
    A[e.to].mo[k] = 1;
    bu.ans[1] = A[e.from].ans[1] - e.mo[k] - e.ans[k];
    bu.mo[1] = A[e.from].mo[1] - e.mo[k];
    for (int t = 1; t < k; t++) {
      bu.ans[t + 1] = A[e.from].ans[t + 1] - e.ans[t];
      bu.mo[t + 1] = A[e.from].mo[t + 1] - e.mo[t];
    }
    A[e.to].ans[1] = e.ans[1] + bu.ans[k] + bu.mo[k];
    A[e.to].mo[1] = e.mo[1] + bu.mo[k];
    for (int t = 1; t < k; t++) {
      A[e.to].ans[t + 1] = e.ans[t + 1] + bu.ans[t];
      A[e.to].mo[t + 1] = e.mo[t + 1] + bu.mo[t];
    }
    for (int i = inof[e.to]; i > -1; i = E[i].next) {
      if (E[i].to != e.from) Q[r++] = i;
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  memset(inof, -1, sizeof(inof));
  long long ans = 0;
  int a, b;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    add(a, b);
  }
  BFS1();
  BFS2();
  for (int i = 1; i <= n; i++)
    for (int t = 1; t <= k; t++) ans += A[i].ans[t];
  cout << ans / 2 << endl;
  return 0;
}

```