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;
int pwr(int base, int p) {
  int ans = 1;
  while (p) {
    if (p & 1) ans = ((ans % 1000000007) * (base % 1000000007)) % 1000000007;
    base = ((base % 1000000007) * (base % 1000000007)) % 1000000007;
    p /= 2;
  }
  return ans;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int powr(int base, int p) {
  int ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base);
    base = (base * base);
    p /= 2;
  }
  return ans;
}
int dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
int dy[] = {0, 0, 1, -1, 1, -1, 1, -1};
set<int> g[100005];
int f[100005];
int main() {
  int n, k;
  scanf("%d", &n);
  scanf("%d", &k);
  for (int i = (0); i < (n - 1); ++i) {
    int x, y;
    scanf("%d", &x);
    scanf("%d", &y);
    g[x].insert(y);
    g[y].insert(x);
  }
  set<int>::iterator it;
  set<int> s;
  for (int i = (0); i < (n); ++i) {
    if (g[i + 1].size() == 1) s.insert(i + 1);
  }
  while (n > 1) {
    set<int> s1;
    for (it = s.begin(); it != s.end(); it++) {
      int tp = *it;
      int u = *(g[tp].begin());
      s1.insert(u);
      g[u].erase(tp);
      f[u]++;
      n--;
    }
    for (it = s1.begin(); it != s1.end(); it++) {
      if (f[*it] < 3) {
        cout << "No";
        return 0;
      }
      f[*it] = 0;
    }
    k--;
    s = s1;
  }
  if (k == 0)
    cout << "Yes";
  else
    cout << "No";
}
