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;
const int N = 1e5 + 5;
vector<int> conn[N];
int degree[N], deleted[N], has[N];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    conn[a].emplace_back(b);
    conn[b].emplace_back(a);
    degree[a]++;
    degree[b]++;
  }
  vector<int> v;
  for (int i = 1; i <= n; ++i)
    if (degree[i] == 1) v.emplace_back(i);
  while (k > 1) {
    k--;
    if (v.size() < 3) return !printf("No\n");
    vector<int> d;
    set<int> st;
    for (const int &a : v) {
      deleted[a] = 1;
      degree[a]--;
      for (const int &b : conn[a]) {
        if (deleted[b]) continue;
        degree[b]--;
        has[b]++;
        st.emplace(b);
      }
    }
    for (const int &b : st) {
      if (degree[b] != 1 or has[b] < 3) return !printf("No\n");
      d.push_back(b);
    }
    v = d;
  }
  int cnt1 = 0, cntx = 0, cnt2 = 0;
  for (int i = 1; i <= n; ++i)
    if (degree[i] == 1)
      cnt1++;
    else if (degree[i] >= 3)
      cntx++;
    else if (degree[i] == 2)
      cnt2++;
  if (cnt2 == 0 and cntx == 1 and cnt1 >= 3)
    puts("Yes");
  else
    puts("No");
}
