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;
const long long inf = 3e12 + 10;
long long n, k;
vector<int> V[1000000];
int deg[1000000];
int d[1000000];
bool dfs(int a, int p, int dist) {
  if (deg[a] == 1) return dist == k;
  bool res = true;
  for (int v : V[a]) {
    if (v != p) {
      res = res && dfs(v, a, dist + 1);
    }
  }
  res = res && (deg[a] >= 4);
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = (0); i < (n - 1); i++) {
    int a, b;
    cin >> a >> b;
    V[a].push_back(b);
    V[b].push_back(a);
    deg[a]++;
    deg[b]++;
    d[a]++;
    d[b]++;
  }
  queue<int> Q;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) Q.push(i);
  }
  while (Q.size() > 1) {
    int a = Q.front();
    Q.pop();
    for (int v : V[a]) {
      d[v]--;
      if (d[v] == 1) {
        Q.push(v);
      }
    }
  }
  int start = Q.front();
  Q.pop();
  deg[start]++;
  if (dfs(start, -1, 0))
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}
