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.
#include <bits/stdc++.h>
using namespace std;
const int MOD9 = 1e9 + 7;
const int MOD91 = 1e9 + 9;
const long long MOD12 = 1e12 + 39LL;
const long long MOD15 = 1e15 + 37LL;
const int INF = 1e9;
const int base = 1e9;
const int MAX = 2e5 + 1;
const long double EPS = 1e-10;
vector<vector<int> > g(MAX);
long long ways[MAX][10], subtree[MAX][10], subAll[MAX], sons[MAX];
int n, k;
int vis[MAX];
void dfs1(int u) {
  vis[u] = 1;
  for (int i = (0); i < (g[u].size()); ++i) {
    int v = g[u][i];
    if (vis[v] == -1) {
      dfs1(v);
      sons[u] += sons[v] + 1;
    }
  }
}
void dfs5(int u) {
  vis[u] = 1;
  for (int i = (0); i < (g[u].size()); ++i) {
    int v = g[u][i];
    if (vis[v] == -1) {
      dfs5(v);
      subAll[u] += subAll[v];
    }
  }
  subAll[u] += sons[u];
}
void dfs(int u) {
  vis[u] = 1;
  for (int i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    if (vis[v] == -1) {
      dfs(v);
      subtree[u][1]++;
      for (int j = (1); j < (k + 1); ++j) {
        subtree[u][j % k] += subtree[v][j - 1];
      }
    }
  }
}
void dfs2(int u) {
  vis[u] = 1;
  int sup[10];
  memset(sup, 0, sizeof(sup));
  for (int i = (0); i < (g[u].size()); ++i) {
    int v = g[u][i];
    if (vis[v] == -1) {
      for (int j = (0); j < (k); ++j) {
        sup[j] += subtree[v][j];
      }
    }
  }
  for (int i = (0); i < (g[u].size()); ++i) {
    int v = g[u][i];
    if (vis[v] == -1) {
      ways[v][1 % k]++;
      ways[v][2 % k] += ((int)g[u].size() - 1) - (u != 1);
      for (int j = (0); j < (k); ++j) {
        sup[j] -= subtree[v][j];
      }
      for (int j = (2); j < (k + 2); ++j) {
        ways[v][j % k] += sup[j - 2];
      }
      for (int j = (1); j < (k + 1); ++j) {
        ways[v][j % k] += ways[u][j - 1];
      }
      dfs2(v);
      for (int j = (0); j < (k); ++j) {
        sup[j] += subtree[v][j];
      }
    }
  }
}
void dfs3(int u) {
  vis[u] = 1;
  for (int i = (0); i < (g[u].size()); ++i) {
    int v = g[u][i];
    if (vis[v] == -1) {
      subAll[v] += subAll[u] - sons[v] - subAll[v] - 1 + (n - sons[v] - 1);
      dfs3(v);
    }
  }
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  memset(vis, -1, sizeof(vis));
  dfs1(1);
  memset(vis, -1, sizeof(vis));
  dfs5(1);
  memset(vis, -1, sizeof(vis));
  dfs(1);
  memset(vis, -1, sizeof(vis));
  dfs2(1);
  memset(vis, -1, sizeof(vis));
  dfs3(1);
  long long sum = 0;
  for (int i = (0); i < (n + 1); ++i) {
    sum += subAll[i];
  }
  for (int i = (1); i < (n + 1); ++i) {
    for (int j = (0); j < (k); ++j) {
      sum += ((ways[i][j] + subtree[i][j]) * ((k - j) % k));
    }
  }
  cout << sum / k / 2;
  return 0;
}
