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;
long long deg[1000005];
vector<int> adj[100005];
int d[1000005];
bool b = 1;
int n, k;
int mx = -1;
void clearq(queue<int> &q) {
  queue<int> empty;
  swap(q, empty);
}
void bfs(int s) {
  queue<int> q;
  d[s] = 0;
  q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (auto v : adj[u]) {
      if (d[v] == -1) d[v] = d[u] + 1, q.push(v), mx = max(mx, d[v]);
    }
  }
  for (int i = 0; i < n; i++) {
    if (adj[i].size() == 1 && d[i] != mx) b = 0;
  }
}
int main() {
  cin >> n >> k;
  memset(d, -1, sizeof d);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    deg[u]++;
    deg[v]++;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  int cnt = 0, c = 0, center;
  queue<int> q1, q2;
  for (int i = 0; i < n; i++) {
    if (deg[i] == 1) q1.push(i);
  }
  while (!q1.empty()) {
    int u = q1.front();
    q1.pop();
    c++;
    for (auto v : adj[u]) {
      deg[v]--;
      if (deg[v] == 1) q2.push(v);
    }
    if (q1.empty() && !q2.empty()) q1 = q2, clearq(q2), cnt++;
    if (c == n) {
      center = u;
    }
  }
  for (int i = 0; i < n && b; i++) {
    if (i != center)
      if (adj[i].size() <= 3 && adj[i].size() > 1) b = 0;
  }
  if (adj[center].size() < 3) b = 0;
  bfs(center);
  cout << (b && c == n && mx == k ? "Yes" : "No") << endl;
  return 0;
}
