Someone give a strange birthday present to Ivan. It is hedgehog — connected undirected graph in which one vertex has degree at least 3 (we will call it center) and all other vertices has degree 1. Ivan thought that hedgehog is too boring and decided to make himself k-multihedgehog.

Let us define k-multihedgehog as follows:

  * 1-multihedgehog is hedgehog: it has one vertex of degree at least 3 and some vertices of degree 1.
  * For all k ≥ 2, k-multihedgehog is (k-1)-multihedgehog in which the following changes has been made for each vertex v with degree 1: let u be its only neighbor; remove vertex v, create a new hedgehog with center at vertex w and connect vertices u and w with an edge. New hedgehogs can differ from each other and the initial gift. 



Thereby k-multihedgehog is a tree. Ivan made k-multihedgehog but he is not sure that he did not make any mistakes. That is why he asked you to check if his tree is indeed k-multihedgehog.

Input

First line of input contains 2 integers n, k (1 ≤ n ≤ 10^{5}, 1 ≤ k ≤ 10^{9}) — number of vertices and hedgehog parameter.

Next n-1 lines contains two integers u v (1 ≤ u,    v ≤ n;    u ≠ v) — indices of vertices connected by edge.

It is guaranteed that given graph is a tree.

Output

Print "Yes" (without quotes), if given graph is k-multihedgehog, and "No" (without quotes) otherwise.

Examples

Input

14 2
1 4
2 4
3 4
4 13
10 5
11 5
12 5
14 5
5 13
6 7
8 6
13 6
9 6


Output

Yes


Input

3 1
1 3
2 3


Output

No

Note

2-multihedgehog from the first example looks like this:

<image>

Its center is vertex 13. Hedgehogs created on last step are: [4 (center), 1, 2, 3], [6 (center), 7, 8, 9], [5 (center), 10, 11, 12, 13].

Tree from second example is not a hedgehog because degree of center should be at least 3.
#include <bits/stdc++.h>
using namespace std;
int n, k, d[100005], f[100005];
bool visit[100005];
vector<int> a[100005];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    d[u]++;
    d[v]++;
    a[u].push_back(v);
    a[v].push_back(u);
  }
  if (k > int(log(n - 1) / log(3))) {
    printf("No");
    return 0;
  }
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (d[i] == 1) {
      visit[i] = true;
      q.push(i);
    }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    d[u]--;
    if (f[u] >= k) {
      printf("No");
      return 0;
    }
    for (int i = 0; i < (int)a[u].size(); i++) {
      int v = a[u][i];
      if (visit[v]) continue;
      if (f[v] == 0)
        if (d[v] == 2)
          if (f[u] < k - 1) {
            printf("No");
            return 0;
          } else
            f[v] = f[u];
        else if (d[v] < 4 && f[u] < k - 1) {
          printf("No");
          return 0;
        } else
          f[v] = f[u] + 1;
      else if (f[v] != f[u] + 1) {
        printf("No");
        return 0;
      }
      d[v]--;
      if (d[v] == 1 && f[u] < k - 1) {
        q.push(v);
        visit[v] = true;
      }
      break;
    }
  }
  for (int i = 1; i <= n; i++)
    if (d[i] > 0) {
      printf("No");
      return 0;
    }
  for (int i = 1; i <= n; i++)
    if (f[i] == k) {
      printf("Yes");
      return 0;
    }
  printf("No");
  return 0;
}
