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;
int n, k, head;
int vis[100001], dist[100001], sig = 1;
vector<vector<int>> tree;
stack<int> track;
void dfs(int n) {
  vis[n] = 1;
  if (sig) track.push(n);
  if (tree[n].size() == 1 && track.size() == 2 * k + 1) sig = 0;
  for (int i = 0; i < tree[n].size(); i++) {
    int v = tree[n][i];
    if (vis[v] == 0) dfs(v);
  }
  if (sig) track.pop();
}
void dfs_check(int n) {
  vis[n] = 1;
  if (!(tree[n].size() == 1 || tree[n].size() >= 4)) sig = 0;
  if (tree[n].size() == 1 && dist[n] != k) sig = 0;
  for (int i = 0; i < tree[n].size(); i++) {
    int v = tree[n][i];
    if (vis[v] == 0) {
      dist[v] = 1 + dist[n];
      dfs_check(v);
    }
  }
}
int main() {
  int i, j;
  int a, b;
  scanf("%d %d", &n, &k);
  tree.resize(n + 1);
  for (i = 0; i < n - 1; i++) {
    scanf("%d %d", &a, &b);
    tree[a].push_back(b);
    tree[b].push_back(a);
  }
  if (n <= 2) {
    printf("No");
    return 0;
  }
  for (i = 1; i <= n; i++) {
    if (tree[i].size() == 1) {
      dfs(i);
      j = track.size() / 2;
      while (j) {
        track.pop();
        j--;
      }
      head = 1;
      if (track.size()) head = track.top();
      memset(vis, 0, sizeof vis);
      sig = 1;
      vis[head] = 1;
      for (j = 0; j < tree[head].size(); j++) {
        dist[tree[head][j]] = 1;
        dfs_check(tree[head][j]);
      }
      if (tree[head].size() < 3) sig = 0;
      break;
    }
  }
  if (sig)
    printf("Yes");
  else
    printf("No");
}
