#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void next_int(T &x) {
  x = 0;
  char c;
  bool neg = false;
  while (!isdigit(c = getchar()))
    if (c == '-') neg = true;
  do x = x * 10 + c - 48;
  while (isdigit(c = getchar()));
  if (neg) x = -x;
}
template <typename T>
inline void write_int(T x, char last = 0) {
  if (x < 0) putchar('-'), x = abs(x);
  char tmp[20];
  int cnt = 0;
  while (x >= 10) tmp[cnt++] = x % 10 + 48, x /= 10;
  tmp[cnt] = x + 48;
  for (int i = (cnt); i >= (0); --i) putchar(tmp[i]);
  if (last) putchar(last);
}
struct data {
  long long sum, cnt;
};
const int N = (int)2e5 + 5;
int n, k, sz[N];
data dp[N][6], dp2[N][6], tmp[N][6];
long long ans;
vector<int> g[N];
void dfs(int u, int p) {
  dp[u][0].cnt = sz[u] = 1;
  for (int i = (0); i < (((int)g[u].size())); ++i) {
    int v = g[u][i];
    if (v == p) continue;
    dfs(v, u);
    sz[u] += sz[v];
    if (k == 1) {
      dp[u][0].sum += (dp[v][0].sum + dp[v][0].cnt),
          dp[u][0].cnt += dp[v][0].cnt;
      tmp[v][0].sum = dp[v][0].sum + dp[v][0].cnt, tmp[v][0].cnt = dp[v][0].cnt;
    } else {
      for (int j = (0); j < (k); ++j) {
        if (!j) {
          dp[u][j].sum += dp[v][k - 1].sum, dp[u][j].cnt += dp[v][k - 1].cnt;
          tmp[v][j].sum = dp[v][k - 1].sum, tmp[v][j].cnt = dp[v][k - 1].cnt;
        } else if (j == 1) {
          dp[u][j].sum += (dp[v][0].sum + dp[v][0].cnt),
              dp[u][j].cnt += dp[v][0].cnt;
          tmp[v][j].sum = dp[v][0].sum + dp[v][0].cnt,
          tmp[v][j].cnt = dp[v][0].cnt;
        } else {
          dp[u][j].sum += dp[v][j - 1].sum, dp[u][j].cnt += dp[v][j - 1].cnt;
          tmp[v][j].sum = dp[v][j - 1].sum, tmp[v][j].cnt = dp[v][j - 1].cnt;
        }
      }
    }
  }
}
void dfs_calc(int u, int p) {
  for (int i = (0); i < (k); ++i) ans += (dp[u][i].sum + dp2[u][i].sum);
  for (int i = (0); i < (((int)g[u].size())); ++i) {
    int v = g[u][i];
    if (v == p) continue;
    for (int j = (0); j < (k); ++j)
      dp[u][j].sum -= tmp[v][j].sum, dp[u][j].cnt -= tmp[v][j].cnt;
    if (k == 1) {
      dp2[v][0].sum +=
          (dp[u][0].sum + dp2[u][0].sum + dp2[u][0].cnt + dp[u][0].cnt);
      dp2[v][0].cnt += (dp[u][0].cnt + dp2[u][0].cnt);
    } else {
      for (int j = (0); j < (k); ++j) {
        if (!j) {
          dp2[v][j].sum += (dp[u][k - 1].sum + dp2[u][k - 1].sum);
          dp2[v][j].cnt += (dp[u][k - 1].cnt + dp2[u][k - 1].cnt);
        } else if (j == 1) {
          dp2[v][j].sum +=
              (dp[u][0].sum + dp2[u][0].sum + dp2[u][0].cnt + dp[u][0].cnt);
          dp2[v][j].cnt += (dp[u][0].cnt + dp2[u][0].cnt);
        } else {
          dp2[v][j].sum += (dp[u][j - 1].sum + dp2[u][j - 1].sum);
          dp2[v][j].cnt += (dp2[u][j - 1].cnt + dp[u][j - 1].cnt);
        }
      }
    }
    for (int j = (0); j < (k); ++j)
      dp[u][j].sum += tmp[v][j].sum, dp[u][j].cnt += tmp[v][j].cnt;
    dfs_calc(v, u);
  }
}
int main() {
  next_int(n), next_int(k);
  for (int i = (1); i < (n); ++i) {
    int u, v;
    next_int(u), next_int(v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  dfs_calc(1, 0);
  write_int(ans / 2LL);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 <typename T>
inline void next_int(T &x) {
  x = 0;
  char c;
  bool neg = false;
  while (!isdigit(c = getchar()))
    if (c == '-') neg = true;
  do x = x * 10 + c - 48;
  while (isdigit(c = getchar()));
  if (neg) x = -x;
}
template <typename T>
inline void write_int(T x, char last = 0) {
  if (x < 0) putchar('-'), x = abs(x);
  char tmp[20];
  int cnt = 0;
  while (x >= 10) tmp[cnt++] = x % 10 + 48, x /= 10;
  tmp[cnt] = x + 48;
  for (int i = (cnt); i >= (0); --i) putchar(tmp[i]);
  if (last) putchar(last);
}
struct data {
  long long sum, cnt;
};
const int N = (int)2e5 + 5;
int n, k, sz[N];
data dp[N][6], dp2[N][6], tmp[N][6];
long long ans;
vector<int> g[N];
void dfs(int u, int p) {
  dp[u][0].cnt = sz[u] = 1;
  for (int i = (0); i < (((int)g[u].size())); ++i) {
    int v = g[u][i];
    if (v == p) continue;
    dfs(v, u);
    sz[u] += sz[v];
    if (k == 1) {
      dp[u][0].sum += (dp[v][0].sum + dp[v][0].cnt),
          dp[u][0].cnt += dp[v][0].cnt;
      tmp[v][0].sum = dp[v][0].sum + dp[v][0].cnt, tmp[v][0].cnt = dp[v][0].cnt;
    } else {
      for (int j = (0); j < (k); ++j) {
        if (!j) {
          dp[u][j].sum += dp[v][k - 1].sum, dp[u][j].cnt += dp[v][k - 1].cnt;
          tmp[v][j].sum = dp[v][k - 1].sum, tmp[v][j].cnt = dp[v][k - 1].cnt;
        } else if (j == 1) {
          dp[u][j].sum += (dp[v][0].sum + dp[v][0].cnt),
              dp[u][j].cnt += dp[v][0].cnt;
          tmp[v][j].sum = dp[v][0].sum + dp[v][0].cnt,
          tmp[v][j].cnt = dp[v][0].cnt;
        } else {
          dp[u][j].sum += dp[v][j - 1].sum, dp[u][j].cnt += dp[v][j - 1].cnt;
          tmp[v][j].sum = dp[v][j - 1].sum, tmp[v][j].cnt = dp[v][j - 1].cnt;
        }
      }
    }
  }
}
void dfs_calc(int u, int p) {
  for (int i = (0); i < (k); ++i) ans += (dp[u][i].sum + dp2[u][i].sum);
  for (int i = (0); i < (((int)g[u].size())); ++i) {
    int v = g[u][i];
    if (v == p) continue;
    for (int j = (0); j < (k); ++j)
      dp[u][j].sum -= tmp[v][j].sum, dp[u][j].cnt -= tmp[v][j].cnt;
    if (k == 1) {
      dp2[v][0].sum +=
          (dp[u][0].sum + dp2[u][0].sum + dp2[u][0].cnt + dp[u][0].cnt);
      dp2[v][0].cnt += (dp[u][0].cnt + dp2[u][0].cnt);
    } else {
      for (int j = (0); j < (k); ++j) {
        if (!j) {
          dp2[v][j].sum += (dp[u][k - 1].sum + dp2[u][k - 1].sum);
          dp2[v][j].cnt += (dp[u][k - 1].cnt + dp2[u][k - 1].cnt);
        } else if (j == 1) {
          dp2[v][j].sum +=
              (dp[u][0].sum + dp2[u][0].sum + dp2[u][0].cnt + dp[u][0].cnt);
          dp2[v][j].cnt += (dp[u][0].cnt + dp2[u][0].cnt);
        } else {
          dp2[v][j].sum += (dp[u][j - 1].sum + dp2[u][j - 1].sum);
          dp2[v][j].cnt += (dp2[u][j - 1].cnt + dp[u][j - 1].cnt);
        }
      }
    }
    for (int j = (0); j < (k); ++j)
      dp[u][j].sum += tmp[v][j].sum, dp[u][j].cnt += tmp[v][j].cnt;
    dfs_calc(v, u);
  }
}
int main() {
  next_int(n), next_int(k);
  for (int i = (1); i < (n); ++i) {
    int u, v;
    next_int(u), next_int(v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  dfs_calc(1, 0);
  write_int(ans / 2LL);
  return 0;
}

```