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, M, i, j, K, a, b, rad, mx, rad2, best;
int dist1[100005], dist2[100005], st[100005], top;
bool viz1[100005], viz2[100005], gas, ok[100005];
vector<int> G[100005];
vector<int>::iterator it;
void dfs1(int nod) {
  vector<int>::iterator it;
  viz1[nod] = true;
  for (it = G[nod].begin(); it != G[nod].end(); it++) {
    if (viz1[*it] == false) {
      dist1[*it] = dist1[nod] + 1;
      if (mx < dist1[*it]) {
        mx = dist1[*it];
        rad = *it;
      }
      dfs1(*it);
    }
  }
}
void dfs2(int nod) {
  vector<int>::iterator it;
  viz2[nod] = true;
  for (it = G[nod].begin(); it != G[nod].end(); it++) {
    if (viz2[*it] == false) {
      dist2[*it] = dist2[nod] + 1;
      best = max(best, dist2[*it]);
      if (best == dist2[*it]) rad2 = *it;
      dfs2(*it);
    }
  }
}
void dfs3(int nod) {
  vector<int>::iterator it;
  ok[nod] = true;
  if (gas == true) return;
  st[++top] = nod;
  if (nod == rad2) {
    gas = true;
    return;
  }
  for (it = G[nod].begin(); it != G[nod].end(); it++) {
    if (ok[*it] == false) dfs3(*it);
    if (gas == true) return;
  }
  top--;
}
bool ESTE(int nod, int prec, int K) {
  if (K == 0 && G[nod].size() == 1) return true;
  if (K == 0 && G[nod].size() > 1) return false;
  if (K > 0 && G[nod].size() == 1) return false;
  if (K > 0 && prec != 0 && G[nod].size() < 4) return false;
  vector<int>::iterator it;
  int X = 0;
  for (it = G[nod].begin(); it != G[nod].end(); it++) {
    if (*it == prec) continue;
    if (ESTE(*it, nod, K - 1) == false) {
      X = 1;
      return false;
    }
  }
  return true;
}
int main() {
  cin >> N >> K;
  for (i = 1; i < N; i++) {
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs1(1);
  dfs2(rad);
  if (best != 2 * K) {
    cout << "No";
    return 0;
  }
  dfs3(rad);
  rad = st[K + 1];
  if (G[rad].size() < 3) {
    cout << "No";
    return 0;
  }
  if (ESTE(rad, 0, K) == true)
    cout << "Yes";
  else
    cout << "No";
  return 0;
}
