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;
void FAST() { ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0); }
int main() {
  FAST();
  long long n, k;
  cin >> n >> k;
  vector<long long> adj[n];
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  long long dp[n];
  for (long long i = 0; i < n; i++) dp[i] = LONG_LONG_MAX;
  queue<pair<long long, long long> > q;
  for (long long i = 0; i < n; i++) {
    if (adj[i].size() == 1) q.push(make_pair(i, 0));
  }
  map<long long, bool> visited;
  long long mx = 0;
  long long cnt = 0;
  while (q.size() != 0) {
    pair<long long, long long> t = q.front();
    q.pop();
    long long ver = t.first;
    long long dist = t.second;
    if (visited[ver] == true) continue;
    long long cnt1 = 0;
    long long cnt2 = 0;
    set<long long> s;
    for (long long i = 0; i < adj[ver].size(); i++) {
      long long adjver = adj[ver][i];
      if (visited[adjver] == true) {
        cnt1++;
        s.insert(dp[adjver]);
      } else
        cnt2++;
    }
    if (s.size() > 1) {
      cout << "NO" << endl;
      return 0;
    }
    if (dist == 0) {
      visited[ver] = true;
      dp[ver] = dist;
      mx = max(mx, dist);
      if (dist == k) cnt++;
    } else if (cnt1 >= 3) {
      visited[ver] = true;
      dp[ver] = dist;
      mx = max(mx, dist);
      if (dist == k) cnt++;
    } else {
      cout << "NO" << endl;
      return 0;
    }
    if (dist != 0 && cnt1 == adj[ver].size()) {
      if (dist == k)
        cout << "YES" << endl;
      else
        cout << "NO" << endl;
      return 0;
    } else if (cnt1 != adj[ver].size() - 1) {
      cout << "NO" << endl;
      return 0;
    }
    for (long long i = 0; i < adj[ver].size(); i++) {
      long long adjver = adj[ver][i];
      if (visited[adjver] == false) q.push(make_pair(adjver, dp[ver] + 1));
    }
  }
  if (mx == k && cnt == 1)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
struct Query {
  long long L, R, idx;
};
bool compare(long long a0, long long a1) { return (a0 > a1); }
long long power(long long x, long long y, long long p) {
  long long ans = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) ans = (ans * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return ans;
}
long long x, y, g;
void extendedEuclid(long long a, long long b) {
  if (b == 0) {
    g = a;
    x = 1;
    y = 0;
  } else {
    extendedEuclid(b, a % b);
    long long temp = x;
    x = y;
    y = temp - (a / b) * y;
  }
}
long long moduloInverse(long long a, long long p) {
  extendedEuclid(a, p);
  if (g == 1 && p != 1)
    return (x + p) % p;
  else
    return -1;
}
