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;
long long Ans, sum, f[200005][6], g[200005][6];
int n, m, cnt;
int size[200005];
int head[200005], to[400005], Next[400005];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void add(int u, int v) {
  to[++cnt] = v;
  Next[cnt] = head[u];
  head[u] = cnt;
}
void Dfs1(int u, int fa) {
  size[u] = 1;
  for (register int i = head[u]; i; i = Next[i]) {
    int v = to[i];
    if (v == fa) continue;
    Dfs1(v, u);
    size[u] += size[v];
  }
  if (u != 1) sum += size[u];
  return;
}
void Dfs2(int u, int fa) {
  Ans += sum;
  for (register int i = head[u]; i; i = Next[i]) {
    int v = to[i];
    if (v == fa) continue;
    size[u] -= size[v];
    sum -= size[v];
    size[v] += size[u];
    sum += size[u];
    Dfs2(v, u);
    size[v] -= size[u];
    sum -= size[u];
    size[u] += size[v];
    sum += size[v];
  }
  return;
}
void DP1(int u, int fa) {
  for (register int i = head[u]; i; i = Next[i]) {
    int v = to[i];
    if (v == fa) continue;
    DP1(v, u);
    for (register int k = 1; k < m; ++k) {
      f[u][(k + 1) % m] += f[v][k];
      g[u][(k + 1) % m] += g[v][k];
    }
    g[u][1] += 1 + g[v][0];
    f[u][1] += 1 + g[v][0] + f[v][0];
  }
  return;
}
void DP2(int u, int fa) {
  for (register int i = 0; i < m; ++i) Ans += f[u][i];
  for (register int i = head[u]; i; i = Next[i]) {
    int v = to[i];
    if (v == fa) continue;
    for (register int k = 1; k < m; ++k) {
      f[u][(k + 1) % m] -= f[v][k];
      g[u][(k + 1) % m] -= g[v][k];
    }
    g[u][1] -= 1 + g[v][0];
    f[u][1] -= 1 + g[v][0] + f[v][0];
    for (register int k = 1; k < m; ++k) {
      f[v][(k + 1) % m] += f[u][k];
      g[v][(k + 1) % m] += g[u][k];
    }
    g[v][1] += 1 + g[u][0];
    f[v][1] += 1 + g[u][0] + f[u][0];
    DP2(v, u);
    g[v][1] -= 1 + g[u][0];
    f[v][1] -= 1 + g[u][0] + f[u][0];
    for (register int k = 1; k < m; ++k) {
      f[v][(k + 1) % m] -= f[u][k];
      g[v][(k + 1) % m] -= g[u][k];
    }
    g[u][1] += 1 + g[v][0];
    f[u][1] += 1 + g[v][0] + f[v][0];
    for (register int k = 1; k < m; ++k) {
      f[u][(k + 1) % m] += f[v][k];
      g[u][(k + 1) % m] += g[v][k];
    }
  }
  return;
}
int main() {
  n = read();
  m = read();
  for (register int i = 2; i <= n; ++i) {
    int u, v;
    u = read();
    v = read();
    add(u, v);
    add(v, u);
  }
  if (m == 1) {
    Dfs1(1, 0);
    Dfs2(1, 0);
  } else {
    DP1(1, 0);
    DP2(1, 0);
  }
  printf("%lld\n", Ans >> 1LL);
  return 0;
}
