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;
using ll = long long;
using ld = long double;
const int maxn = 100100;
const int inf = 1e9;
vector<int> g[maxn];
vector<int> from;
vector<int> d;
int n;
int bfs(int u) {
  d.assign(n, inf);
  from.assign(n, -1);
  d[u] = 0;
  queue<int> o;
  o.push(u);
  while (!o.empty()) {
    u = o.front();
    o.pop();
    for (int v : g[u]) {
      if (d[v] == inf) {
        d[v] = d[u] + 1;
        from[v] = u;
        o.push(v);
      }
    }
  }
  return u;
}
bool check(int u, int k) {
  if (k == 0) {
    return g[u].empty();
  }
  if ((int)g[u].size() < 3) {
    return false;
  }
  for (int v : g[u]) {
    auto it = find(g[v].begin(), g[v].end(), u);
    assert(it != g[v].end());
    g[v].erase(it);
    if (!check(v, k - 1)) {
      return false;
    }
  }
  return true;
}
signed main() {
  int k;
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int u = bfs(0);
  int v = bfs(u);
  if (d[v] != 2 * k) {
    cout << "No\n";
    return 0;
  }
  for (int i = 0; i < k; ++i) {
    v = from[v];
  }
  cout << (check(v, k) ? "Yes" : "No") << '\n';
}
