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 k, n, root, par[100005];
vector<int> v[100005];
bool vis[100005], fr;
void findroot(int u, int l) {
  int i;
  if (v[u].size() == 1 && l == 2 * k) {
    fr = true;
    root = u;
  }
  for (i = 0; i < v[u].size(); i++) {
    if (fr) break;
    if (!vis[v[u][i]]) {
      vis[v[u][i]] = true;
      par[v[u][i]] = u;
      findroot(v[u][i], l + 1);
    }
  }
}
void dfs(int u, int l) {
  int i;
  if (v[u].size() == 1 && l != k) fr = true;
  if (u != root && l != k && v[u].size() < 4) fr = true;
  for (i = 0; i < v[u].size(); i++) {
    if (fr) break;
    if (!vis[v[u][i]]) {
      vis[v[u][i]] = true;
      dfs(v[u][i], l + 1);
    }
  }
}
int main() {
  int i, x, y, leaf, cnt = 0;
  cin >> n >> k;
  for (i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  if (n == 1) {
    cout << "No\n";
    return 0;
  }
  for (i = 1; i <= n; i++)
    if (v[i].size() == 1) {
      leaf = i;
      break;
    }
  memset(vis, false, sizeof(vis));
  vis[leaf] = true;
  fr = false;
  findroot(leaf, 0);
  if (!fr) {
    cout << "No\n";
    return 0;
  }
  while (cnt != k) {
    root = par[root];
    cnt++;
  }
  if (v[root].size() < 3) {
    cout << "No\n";
    return 0;
  }
  memset(vis, false, sizeof(vis));
  vis[root] = true;
  fr = false;
  dfs(root, 0);
  if (fr) {
    cout << "No\n";
  } else
    cout << "Yes\n";
  return 0;
}
