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 double PI = 2 * acos(0.0);
const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";
template <class T>
ostream &operator<<(ostream &out, vector<T> &a) {
  out << "(";
  for (auto x : a) out << x << " ";
  return out << ")";
}
void smain();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  smain();
}
const int N = 1e5 + 9;
int n, k;
vector<int> g[N];
int dp[N];
void dfs1(int v, int par) {
  for (int to : g[v])
    if (to != par) {
      dfs1(to, v);
      dp[v] = max(dp[v], (dp[to] + 1));
    }
}
void dfs2(int v, int par, int way, int &root) {
  if (way == 0) {
    root = v;
    return;
  }
  int go = v;
  int val = -1;
  for (int to : g[v])
    if (to != par) {
      if (dp[to] > val) {
        val = dp[to];
        go = to;
      }
    }
  assert(go != v);
  dfs2(go, v, way - 1, root);
}
bool ok = 1;
void dfs3(int v, int par, int depth) {
  if (g[v].size() == 1) {
    ok &= depth == k;
    return;
  }
  int ch = 0;
  for (int to : g[v])
    if (to != par) {
      ch++;
      dfs3(to, v, depth + 1);
    }
  ok &= ch >= 3;
}
void smain() {
  cin >> n >> k;
  for (int i = 0; i < int(n - 1); ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int tmp_root = 1;
  for (int i = 1; i <= n; ++i)
    if (g[i].size() == 1) tmp_root = i;
  dfs1(tmp_root, -1);
  int way = dp[tmp_root];
  if (way & 1) {
    cout << "No";
    return;
  }
  way >>= 1;
  int root = tmp_root;
  dfs2(tmp_root, -1, way, root);
  dfs3(root, -1, 0);
  if (ok)
    cout << "Yes";
  else
    cout << "No";
}
