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;
vector<long long int> tree[100010];
long long int dist[100010];
long long int vertice, k;
bool visit[100010];
long long int parent[100010];
bool DFS(long long int start, long long int par, long long int k) {
  long long int size, store, count = 0;
  bool ans = true, res;
  size = tree[start].size();
  for (long long int i = 0; i < size; i++) {
    store = tree[start][i];
    if (store != par) {
      res = DFS(store, start, k - 1);
      count++;
      ans = ans && res;
    }
  }
  if (count == 0) {
    if (k == 0) {
      return true;
    }
    return false;
  }
  if (count < 3) {
    return false;
  }
  return ans;
}
long long int bfs(long long int vertex) {
  queue<long long int> q1;
  for (long long int i = 1; i <= vertice; i++) {
    visit[i] = false;
    dist[i] = 0;
  }
  q1.push(vertex);
  visit[vertex] = true;
  long long int size, store, front;
  while (!q1.empty()) {
    front = q1.front();
    q1.pop();
    size = tree[front].size();
    for (long long int i = 0; i < size; i++) {
      store = tree[front][i];
      if (!visit[store]) {
        visit[store] = true;
        parent[store] = front;
        dist[store] = dist[front] + 1;
        q1.push(store);
      }
    }
  }
  return front;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> vertice >> k;
  long long int vert1, vert2;
  for (long long int i = 1; i < vertice; i++) {
    cin >> vert1 >> vert2;
    tree[vert1].push_back(vert2);
    tree[vert2].push_back(vert1);
  }
  long long int diam1 = bfs(1LL);
  long long int diam2 = bfs(diam1);
  long long int half = dist[diam2];
  if (half % 2 == 1) {
    cout << "No";
    exit(0);
  }
  half = half / 2;
  long long int centre;
  while (half != 0) {
    diam2 = parent[diam2];
    half--;
  }
  centre = diam2;
  bool res1 = DFS(centre, -1, k);
  if (res1) {
    cout << "Yes";
    exit(0);
  }
  cout << "No";
  return 0;
}
