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>
#pragma GCC optimize(3)
using namespace std;
int n, k;
vector<int> G[100005];
int deg[100005];
vector<int> v;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
    deg[v]++;
    deg[u]++;
  }
  if (n == 1) {
    puts("No");
    return 0;
  }
  int c = 0, id = -1;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 2) {
      puts("No");
      return 0;
    }
    if (deg[i] == 3) {
      c++;
      id = i;
    }
  }
  if (c >= 2) {
    puts("No");
    return 0;
  }
  int cur = 0, del = 0;
  while (del < n) {
    v.clear();
    for (int i = 1; i <= n; i++)
      if (deg[i] == 1) v.push_back(i);
    if (del == n - 1) {
      if (c == 1) {
        if (v[0] != id) {
          puts("No");
          return 0;
        }
      }
    }
    set<int> s;
    for (auto it : v) {
      del++;
      for (auto id : G[it])
        if (deg[id] > 1) {
          s.insert(id);
          deg[id]--;
        }
      deg[it]--;
    }
    for (auto it : s)
      if (deg[it] != 1) {
        puts("No");
        return 0;
      }
    for (int i = 1; i <= n; i++)
      if (deg[i] == 2) {
        puts("No");
        return 0;
      }
    cur++;
    if (del == n) {
      if ((int)v.size() != 1) {
        puts("No");
        return 0;
      }
    }
  }
  if (cur == k + 1)
    puts("Yes");
  else
    puts("No");
  return 0;
}
