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("-O2")
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
const int LIM = 1e5 + 5, MOD = 1e9 + 7;
int n, k;
std::vector<vector<int> > graph;
vector<int> firstdistance;
std::vector<int> seconddistance;
int dfs(const int &prev, const int &curr, int pathlen) {
  int siz = graph[curr].size();
  firstdistance[curr] = 2 * k - pathlen;
  if (pathlen == 0) {
    if (siz == 1) return curr;
    return -1;
  }
  for (int i = 0; i < siz; i++) {
    if (graph[curr][i] != prev) {
      int root = dfs(curr, graph[curr][i], pathlen - 1);
      if (root != -1) return root;
    }
  }
  return -1;
}
void distance2(const int &prev, const int &curr, int pathlen) {
  int siz = graph[curr].size();
  seconddistance[curr] = pathlen;
  for (int i = 0; i < siz; i++) {
    if (graph[curr][i] != prev) {
      distance2(curr, graph[curr][i], pathlen + 1);
    }
  }
}
bool dfs2(const int &prev, const int &curr, int pathlen) {
  int siz = graph[curr].size();
  if (siz == 1) {
    if (k == pathlen) {
      return true;
    } else
      return false;
  } else if (k == pathlen) {
    return false;
  } else if (siz <= 3) {
    if (pathlen != 0)
      return false;
    else if (siz <= 2)
      return false;
  }
  if (pathlen > k) return false;
  for (int i = 0; i < siz; i++) {
    if (graph[curr][i] != prev) {
      bool final = dfs2(curr, graph[curr][i], pathlen + 1);
      if (!final) return false;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  graph.assign(n + 1, std::vector<int>());
  firstdistance.assign(n + 1, -1);
  seconddistance.assign(n + 1, -2);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  if (k > n || n <= 3) {
    cout << "No";
    return 0;
  }
  int lf = 0;
  for (int i = 1; i < n + 1; i++) {
    if (graph[i].size() == 1) {
      lf = i;
      break;
    }
  }
  int otherend = dfs(-1, lf, 2 * k);
  if (otherend == -1) {
    cout << "No" << endl;
    return 0;
  }
  distance2(-1, otherend, 0);
  int cent = -1;
  for (int i = 1; i < n + 1; i++) {
    if (firstdistance[i] == k && seconddistance[i] == k) {
      cent = i;
      break;
    }
  }
  if (cent == -1) {
    cout << "No";
    return 0;
  }
  if (dfs2(-1, cent, 0))
    cout << "Yes";
  else
    cout << "No";
}
