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 maxn = 100010;
vector<int> g[maxn];
set<int> g_dic[maxn];
int dep[maxn], k;
bool ok = true;
void dfs(int u, int fa) {
  if (dep[u] == k) {
    if (g[u].size() != 1) ok = false;
  } else {
    if (g[u].size() - (fa == -1 ? 0 : 1) < 3) ok = false;
  }
  for (int i = 0; i < (int)g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    dep[v] = dep[u] + 1;
    dfs(v, u);
  }
}
vector<int> ord;
int main() {
  int n, u, v;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    g_dic[u].insert(v);
    g_dic[v].insert(u);
  }
  if (n == 1) {
    puts("No");
    exit(0);
  }
  queue<int> q;
  for (int v = 1; v <= n; v++)
    if ((int)g_dic[v].size() == 1) q.push(v);
  while (!q.empty()) {
    int v = q.front();
    ord.push_back(v);
    q.pop();
    if (g_dic[v].size()) {
      int par = *g_dic[v].begin();
      g[par].push_back(v);
      g[v].push_back(par);
      g_dic[par].erase(v);
      g_dic[v].erase(par);
      if ((int)g_dic[par].size() <= 1) q.push(par);
    }
  }
  ok = true;
  dfs(ord.back(), -1);
  bool ok1 = ok;
  ok = true;
  dfs(ord[(int)ord.size() - 2], -1);
  bool ok2 = ok;
  puts(ok1 || ok2 ? "Yes" : "No");
  return 0;
}
