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 + 10;
vector<int> adj[N];
int n, k, tar, key, p[N], dp[N];
void dfs(int u, int fa = 0, int dep = 0) {
  p[u] = fa;
  dp[u] = dep;
  if (dep > key) key = dep, tar = u;
  for (auto v : adj[u])
    if (v != fa) {
      dfs(v, u, dep + 1);
    }
}
int find(int u) {
  tar = u, key = 0;
  dfs(u);
  return tar;
}
int main() {
  cin >> n >> k;
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int st, ed, cent;
  st = find(1);
  ed = find(st);
  if (dp[ed] & 1) return puts("No"), 0;
  cent = ed;
  while (dp[cent] > dp[ed] / 2) cent = p[cent];
  ed = find(cent);
  if (dp[ed] != k) return puts("No"), 0;
  for (int i = 1; i <= n; i++) {
    if (i == cent) {
      if (adj[i].size() < 3) return puts("No"), 0;
    } else if (adj[i].size() == 1) {
      if (dp[i] != dp[ed]) return puts("No"), 0;
    } else {
      if (dp[i] >= dp[ed]) return puts("No"), 0;
      if (adj[i].size() < 4) return puts("No"), 0;
    }
  }
  puts("Yes");
  return 0;
}
