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;
vector<int> E[100005];
int d[100005];
int n, k, tar;
int bfs(int s) {
  for (int i = 1; i <= n; i++) d[i] = -1;
  d[s] = 0;
  queue<int> q;
  q.push(s);
  while (q.size()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < E[u].size(); i++) {
      if (d[E[u][i]] == -1) {
        d[E[u][i]] = d[u] + 1;
        q.push(E[u][i]);
      }
    }
  }
  int mx = 1;
  for (int i = 2; i <= n; i++) {
    if (d[i] > d[mx]) mx = i;
  }
  return mx;
}
vector<int> pth;
bool dfs(int fa, int u) {
  for (int i = 0; i < E[u].size(); i++) {
    if (E[u][i] == tar) {
      pth.push_back(E[u][i]);
      pth.push_back(u);
      return 1;
    } else if (E[u][i] != fa) {
      bool q = dfs(u, E[u][i]);
      if (q) {
        pth.push_back(u);
        return 1;
      }
    }
  }
  return 0;
}
void ddfs(int fa, int u, int d) {
  if (E[u].size() > 1 && E[u].size() < 4 && fa) {
    puts("No");
    exit(0);
  } else if (fa == 0 && E[u].size() < 3) {
    puts("No");
    exit(0);
  } else if (E[u].size() == 1 && d != k) {
    puts("No");
    exit(0);
  }
  for (int i = 0; i < E[u].size(); i++) {
    if (E[u][i] != fa) ddfs(u, E[u][i], d + 1);
  }
  return;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 2; i <= n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    E[u].push_back(v);
    E[v].push_back(u);
  }
  int s = bfs(1);
  int t = bfs(s);
  tar = t;
  if (k * 2 != d[t]) {
    puts("No");
    return 0;
  }
  dfs(0, s);
  ddfs(0, pth[k], 0);
  puts("Yes");
  return 0;
}
