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;
class Adj {
 public:
  set<int> list;
  int id;
};
int main() {
  int n, k;
  cin >> n >> k;
  if (k > 13) {
    cout << "No" << endl;
    return 0;
  }
  Adj adj[n + 1];
  adj[n].id = n;
  for (int x, y, i = 1; i < n; i++) {
    adj[i].id = i;
    cin >> x >> y;
    adj[x].list.insert(y);
    adj[y].list.insert(x);
  }
  int i = 1, depth = 0;
  bool flag = true;
  while (i < n) {
    sort(adj + i, adj + n + 1,
         [](Adj &lhs, Adj &rhs) { return lhs.list.size() < rhs.list.size(); });
    map<int, int> ref;
    for (int j = i; j <= n; j++) {
      ref[adj[j].id] = j;
    }
    int j = i;
    while (adj[i].list.size() == 1) i++;
    map<int, int> mp;
    for (; j < i; j++) {
      int x = adj[j].id;
      int y = *(adj[j].list.begin());
      mp[y]++;
      adj[ref[y]].list.erase(x);
    }
    map<int, int>::iterator it;
    for (it = mp.begin(); it != mp.end(); it++) {
      if (it->second < 3) {
        flag = false;
        break;
      }
    }
    if (!flag) break;
    depth++;
  }
  if (flag && depth == k && i == n)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}
