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>
#pragma comment(linker, "/STACK:10000000000")
using namespace std;
const int MOD = 1000000007;
const int INF = 1000000007LL;
const long long INF2 = 1000000007LL * 1000000007LL;
const long double EPS = 1e-9;
const int SIZE = 100100;
mt19937 rng(time(0));
uniform_int_distribution<int> uid(-1000000000, 1000000000);
int n, m, k, cnt, n2;
set<int> g[SIZE];
int u[SIZE];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int rep_iter = 0; rep_iter < n - 1; ++rep_iter) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    g[a].insert(b);
    g[b].insert(a);
  }
  set<int> cur, nxt;
  n2 = n;
  for (int i = 0; i < n; ++i) {
    if (g[i].size() == 1) {
      cur.insert(i);
    }
  }
  while (n > 1) {
    for (auto& i : cur) {
      int v = *g[i].begin();
      if ((cur.find(v) != cur.end())) {
        cout << "No";
        return 0;
      }
      nxt.insert(v);
      u[v]++;
      g[v].erase(i);
    }
    ++cnt;
    n -= cur.size();
    for (auto& i : nxt) {
      if (g[i].size() != 1 && nxt.size() > 1 ||
          g[i].size() != 0 && nxt.size() == 1 || u[i] < 3) {
        cout << "No";
        return 0;
      }
    }
    cur = nxt;
    nxt.clear();
  }
  if (cnt == k)
    cout << "Yes";
  else
    cout << "No";
  return 0;
}
