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 N = 100005;
vector<int> adj[N];
int d[N], del[N], l[N];
int main() {
  memset(l, -1, sizeof l);
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[--a].push_back(--b);
    adj[b].push_back(a);
    ++d[a], ++d[b];
  }
  queue<int> q, nq;
  for (int i = 0; i < n; ++i)
    if (d[i] == 1) q.push(i), l[i] = 0;
  int cnt = 0;
  while (q.size() > 1) {
    while (!q.empty()) {
      int u = q.front();
      if (adj[u].size() != 1 && adj[u].size() <= 3) {
        printf("No\n");
        return 0;
      }
      q.pop();
      for (int v : adj[u]) {
        if (l[v] == -1)
          l[v] = l[u] + 1;
        else if (l[v] != l[u] + 1 && d[v] > 1) {
          printf("No\n");
          return 0;
        }
        if (--d[v] == 1) nq.push(v);
      }
    }
    q = nq;
    while (!nq.empty()) nq.pop();
    ++cnt;
  }
  if (q.size() == 1 && cnt == k && adj[q.front()].size() >= 3)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
