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 MOD1 = 1e9 + 7;
const int MOD2 = 998244353;
const long long INF = 2 * 1e18;
const long double PI = 3.14159265358979323846;
vector<long long> ed[100005];
long long dist[100005][5], sz[100005];
void dfs1(long long n, long long p, long long x) {
  for (long long i = 0; i < ed[n].size(); i++) {
    if (ed[n][i] == p) continue;
    dist[ed[n][i]][x] = dist[n][x] + 1;
    dfs1(ed[n][i], n, x);
  }
}
void dfs2(long long n, long long p) {
  long long cnt = 0;
  for (long long i = 0; i < ed[n].size(); i++) {
    if (ed[n][i] != p) {
      dist[ed[n][i]][2] = dist[n][2] + 1;
      cnt++;
      dfs2(ed[n][i], n);
    }
  }
  sz[n] = cnt;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k, pos = -1;
  bool x = 0;
  cin >> n >> k;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    ed[u].push_back(v);
    ed[v].push_back(u);
  }
  for (long long i = 1; i <= n; i++) {
    if (ed[i].size() == 1) {
      dfs1(i, -1, 0);
      break;
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (dist[i][0] == 2 * k) {
      pos = i;
      x = 1;
    }
    if (dist[i][0] > 2 * k) {
      cout << "No";
      return 0;
    }
  }
  if (!x) {
    cout << "No";
    return 0;
  }
  dfs1(pos, -1, 1);
  x = 0;
  for (long long i = 1; i <= n; i++) {
    if (dist[i][0] == k && dist[i][1] == k) {
      dfs2(i, -1);
      x = 1;
      break;
    }
  }
  if (!x) {
    cout << "No";
    return 0;
  }
  x = 0;
  for (long long i = 1; i <= n; i++) {
    if (dist[i][2] == k) {
      if (ed[i].size() != 1) x = 1;
    } else {
      if (sz[i] < 3) x = 1;
    }
  }
  if (!x)
    cout << "Yes";
  else
    cout << "No";
}
