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;
long long n, k, r, second, p[200005], xx, yy, ch[200005];
pair<long long, long long> d[200005];
vector<long long> v[100005];
void go(long long x, long long par) {
  p[x] = par;
  if (v[x].size() == 1 && v[x][0] == par) {
    d[x].first = 0;
    d[x].second = x;
  }
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i] != par) {
      go(v[x][i], x);
      if (d[v[x][i]].first + 1 >= d[x].first) {
        d[x].first = d[v[x][i]].first + 1;
        d[x].second = d[v[x][i]].second;
      }
    }
  }
}
void dfs(long long x, long long par, long long dep) {
  if (v[x].size() == 1 && v[x][0] == par) {
    if (dep != k) {
      cout << "No";
      exit(0);
    }
    return;
  }
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i] != par) {
      dfs(v[x][i], x, dep + 1);
      ch[x]++;
    }
  }
  if (ch[x] < 3) {
    cout << "No";
    exit(0);
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    cin >> xx >> yy;
    v[xx].push_back(yy);
    v[yy].push_back(xx);
  }
  go(1, -1);
  if (d[1].first < k) {
    cout << "No";
    return 0;
  }
  second = d[1].second;
  r = 0;
  while (r < k) {
    second = p[second];
    r++;
  }
  dfs(second, -1, 0);
  cout << "Yes";
  return 0;
}
