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 long long MOD = 1000000007;
const long long INF = 1000000000;
template <class T>
T read(T &x) {
  char c;
  int flag = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag *= -1;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') x = x * 10 + c - '0';
  x *= flag;
  return x;
}
int dep[100005];
int fa[100005];
vector<int> adj[100005];
void dfs(int x, int f) {
  for (int i : adj[x]) {
    if (i == f) continue;
    fa[i] = x;
    dep[i] = dep[x] + 1;
    dfs(i, x);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, -1);
  int root = 1;
  for (int i = 1; i <= n; ++i)
    if (dep[root] < dep[i]) root = i;
  memset(dep, 0, sizeof dep);
  ;
  dfs(root, -1);
  int tar = root;
  for (int i = 1; i <= n; ++i)
    if (dep[tar] < dep[i]) tar = i;
  vector<int> v;
  while (tar != root) {
    v.push_back(tar);
    tar = fa[tar];
  }
  v.push_back(root);
  if (v.size() != (k << 1) + 1) return cout << "No" << endl, 0;
  memset(dep, 0, sizeof dep);
  ;
  root = v[k];
  dfs(root, -1);
  for (int i = 1; i <= n; ++i) {
    if (dep[i] == k) continue;
    int sz = adj[i].size() - 1 + (root == i);
    if (sz < 3) return cout << "No" << endl, 0;
  }
  cout << "Yes" << endl;
}
