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>
const double PI = 3.1415926535897932384626433;
const int KL = 1e5;
const long long MOD = 998244353;
using namespace std;
long long n, a[KL], k, x, y, len[KL], pr[KL], z;
bool vis[KL];
vector<long long> adj[KL];
vector<long long> vec, path;
long long pw[1000];
void dfs(long long node, long long p, long long dis) {
  len[node] = dis;
  pr[node] = p;
  for (auto v : adj[node]) {
    if (v == p) continue;
    dfs(v, node, dis + 1);
  }
}
int main() {
  pw[0] = 1;
  for (long long i = 1; i <= 11; i++) {
    pw[i] = pw[i - 1] * 3;
  }
  scanf("%lld", &n);
  scanf("%lld", &k);
  long long e = n - 1;
  while (e--) {
    scanf("%lld", &x);
    scanf("%lld", &y);
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  if (k >= 11) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() == 1) {
      vec.push_back(i);
    }
  }
  if (vec.size() < pw[k]) {
    cout << "NO" << endl;
    return 0;
  }
  x = vec[0];
  dfs(x, 0, 0);
  bool flag = false;
  for (int i = 0; i < vec.size(); i++) {
    if (len[vec[i]] == 2 * k) {
      flag = true;
      y = vec[i];
      break;
    }
  }
  if (!flag) {
    cout << "NO" << endl;
    return 0;
  }
  int cnt = 0;
  while (true) {
    y = pr[y];
    cnt++;
    if (cnt == k) {
      z = y;
      break;
    }
  }
  dfs(z, 0, 0);
  bool ok = (adj[z].size() >= 3);
  for (int i = 0; i < vec.size(); i++) {
    if (len[vec[i]] != k) ok = false;
    vis[vec[i]] = 1;
  }
  for (int i = 1; i <= n; i++) {
    if (i == z) continue;
    if (!vis[i] && adj[i].size() <= 3) ok = false;
  }
  if (!ok) {
    cout << "NO" << endl;
  } else
    cout << "YES" << endl;
  return 0;
}
