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, ki, h, hh;
int d[100001], dg[100001], apa[100001];
bool volt[100001];
vector<int> g[100001];
bool good = true;
void mely(int a) {
  volt[a] = true;
  for (int i = 0; i < g[a].size(); i++) {
    if (volt[g[a][i]]) continue;
    d[g[a][i]] = d[a] + 1;
    apa[g[a][i]] = a;
    mely(g[a][i]);
  }
}
int md(int a) {
  for (int i = 1; i <= n; i++) volt[i] = false;
  d[a] = 0;
  apa[a] = a;
  mely(a);
  int maxd = 1;
  for (int i = 2; i <= n; i++) {
    if (d[i] > d[maxd]) maxd = i;
  }
  return maxd;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  if (k > n) k = n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
    dg[a]++;
    dg[b]++;
  }
  for (int i = 1; i <= n; i++) {
    if (dg[i] == 2) good = false;
    if (dg[i] == 3) {
      h++;
      hh = i;
    }
  }
  if (h > 1) good = false;
  if (h == 0) {
    int a = md(1);
    int b = md(a);
    for (int i = 0; i < k; i++) b = apa[b];
    hh = b;
  }
  if (h <= 1) {
    int a = md(hh);
    if (d[a] != k) good = false;
    for (int i = 1; i <= n; i++) {
      if (dg[i] > 1) continue;
      if (d[i] == k) continue;
      good = false;
    }
  }
  if (good)
    cout << "Yes\n";
  else
    cout << "No\n";
  return 0;
}
