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 MAXN = 100005;
const int MOD = 998244353;
int N, K;
vector<int> adj[MAXN];
bool vis[MAXN];
pair<long long, long long> prevnode[MAXN];
pair<long long, long long> dfs1(long long v) {
  pair<long long, long long> res(0, v);
  vis[v] = 1;
  for (auto i : adj[v]) {
    if (vis[i]) continue;
    prevnode[i] = pair<long long, long long>(v, 1);
    pair<long long, long long> cur = dfs1(i);
    cur.first++;
    res = max(res, cur);
  }
  vis[v] = 0;
  return res;
}
bool poss = true;
int findDiameter(long long v0) {
  long long v1 = dfs1(v0).second;
  pair<long long, long long> tmp = dfs1(v1);
  long long totaldist = tmp.first, v2 = tmp.second;
  pair<long long, long long> prv(-1, totaldist), cur(0, totaldist);
  long long curnode = v2;
  while (curnode != v1) {
    prv = cur;
    cur.first += prevnode[curnode].second;
    cur.second -= prevnode[curnode].second;
    curnode = prevnode[curnode].first;
    if (cur.first == cur.second) {
      return curnode;
    } else if (cur.first > cur.second) {
      poss = false;
      return curnode;
    }
  }
  assert(false);
  return curnode;
}
int rt = 1;
int mx = 0;
void dfs(int v, int h) {
  if (vis[v]) return;
  vis[v] = true;
  mx = max(mx, h);
  if (h != K && ((int)(adj[v]).size()) == 1) {
    poss = false;
  }
  if (v == rt) {
    if (((int)(adj[v]).size()) < 3) poss = false;
  } else {
    if (((int)(adj[v]).size()) != 1 && ((int)(adj[v]).size()) < 4) poss = false;
  }
  for (auto i : adj[v]) {
    if (vis[i]) continue;
    dfs(i, h + 1);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> K;
  if (N <= 3) {
    cout << "No\n";
    return 0;
  }
  for (int i = (1); i <= (N - 1); i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  rt = findDiameter(1);
  dfs(rt, 0);
  if (mx != K) poss = false;
  if (poss)
    cout << "Yes\n";
  else
    cout << "No\n";
}
