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;
using ll = long long;
using vll = vector<ll>;
vector<vector<long long>> adjlist;
long long dfs(long long d, long long cur, long long pref = -1) {
  if (d == 0) {
    return cur;
  } else {
    for (long long n : adjlist[cur]) {
      if (n != pref) {
        long long res = dfs(d - 1, n, cur);
        if (res >= 0) return res;
      }
    }
    return -1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, k;
  cin >> n >> k;
  if (n == 1) {
    cout << "No" << endl;
    return 0;
  }
  adjlist.assign(n, {});
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    adjlist[a].push_back(b);
    adjlist[b].push_back(a);
  }
  vector<long long> level;
  vector<bool> seen(n);
  for (long long i = 0; i < n; i++) {
    if (adjlist[i].size() == 1) {
      level.push_back(i);
      seen[i] = true;
    }
  }
  for (long long i = 0; i < k && !level.empty(); i++) {
    vector<long long> nextLevel;
    for (long long j : level) {
      for (long long k : adjlist[j]) {
        if (!seen[k]) {
          seen[k] = true;
          nextLevel.push_back(k);
        }
      }
    }
    level = nextLevel;
  }
  if (level.size() != 1 || adjlist[level[0]].size() < 3) {
    cout << "No" << endl;
    return 0;
  }
  seen.assign(n, false);
  seen[level[0]] = true;
  level = adjlist[level[0]];
  for (long long i = 1; i < k; i++) {
    vector<long long> nextLevel;
    for (long long x : level) {
      seen[x] = true;
      if (adjlist[x].size() <= 3) {
        cout << "No" << endl;
        return 0;
      }
      for (long long n : adjlist[x]) {
        if (!seen[n]) {
          nextLevel.push_back(n);
        }
      }
    }
    level = nextLevel;
  }
  for (long long i : level) {
    if (adjlist[i].size() != 1) {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
}
