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("Ofast", "unroll-loops")
using namespace std;
template <class T>
inline void sort(T &a) {
  sort((a).begin(), (a).end());
}
template <class T>
inline void rsort(T &a) {
  sort((a).rbegin(), (a).rend());
}
template <class T>
inline void reverse(T &a) {
  reverse((a).begin(), (a).end());
}
template <class T, class U>
inline void checkmin(T &x, U y) {
  if (x > y) x = y;
}
template <class T, class U>
inline void checkmax(T &x, U y) {
  if (x < y) x = y;
}
vector<vector<int>> G;
int head;
long long n, k;
bool dfscheck(int v, int p, int d) {
  if (G[v].size() == 1) return d == k;
  if (G[v].size() < 3 + (p != -1)) return false;
  for (auto i : G[v])
    if (i != p)
      if (!dfscheck(i, v, d + 1)) return false;
  return true;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(20);
  srand(time(NULL));
  cin >> n >> k;
  if (n < 4) {
    cout << "No";
    return 0;
  }
  G = vector<vector<int>>(n);
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  deque<int> que;
  vector<bool> used(n);
  vector<int> near(n);
  for (int i = 0; i < n; ++i)
    if (int(G[i].size()) == 1) que.push_back(i), used[i] = true, near[i] = 0;
  while (int(que.size()) > 0) {
    int fr = que.front();
    que.pop_front();
    for (auto i : G[fr])
      if (!used[i]) {
        used[i] = true;
        near[i] = near[fr] + 1;
        que.push_back(i);
      }
  }
  int ind = 0;
  for (int i = 0; i < n; ++i)
    if (near[i] > near[ind]) ind = i;
  if (near[ind] != k) {
    cout << "No";
    return 0;
  }
  if (dfscheck(ind, -1, 0))
    cout << "Yes";
  else
    cout << "No";
  return 0;
}
