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 int mx = 1e5 + 5;
vector<int> adj[mx];
int deg[mx], vis[mx], dist[mx], root = 1;
void precomp(int s, int p) {
  dist[s] = dist[p] + 1;
  for (int t : adj[s]) {
    if (t == p) continue;
    precomp(t, s);
  }
}
int check(int s, int p) {
  if (deg[s] == 1) return 1;
  int cnt = 0, ret = 1;
  for (int t : adj[s]) {
    if (t == p) continue;
    ret &= check(t, s);
    ++cnt;
  }
  if (cnt < 3) ret &= 0;
  return ret;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
    ++deg[a], ++deg[b];
  }
  queue<int> q;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 1) q.push(i);
  }
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    if (vis[cur]) continue;
    vis[cur] = 1, root = cur;
    for (int p : adj[cur]) q.push(p);
  }
  precomp(root, 0);
  for (int i = 1; i <= n; ++i) {
    if ((deg[i] == 1) && (dist[i] != (k + 1))) {
      puts("No");
      return 0;
    }
  }
  if (check(root, 0))
    puts("Yes");
  else
    puts("No");
  return 0;
}
