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 long long N = 1e5 + 10, LG = 30, mod = 998244353;
std::vector<long long> g[N];
long long pred[N];
long long n;
std::vector<long long> bfs(long long start) {
  std::queue<long long> q;
  std::vector<long long> d(n, 1e9);
  d[start] = 0;
  q.push(start);
  while (!q.empty()) {
    long long v = q.front();
    q.pop();
    for (auto e : g[v]) {
      if (d[e] > d[v] + 1) {
        d[e] = d[v] + 1;
        q.push(e);
      }
    }
  }
  return d;
}
void dfs(long long v, long long p = -1) {
  pred[v] = p;
  for (auto e : g[v]) {
    if (e != p) {
      dfs(e, v);
    }
  }
}
int32_t main() {
  std::ios::sync_with_stdio(false);
  long long k;
  std::cin >> n >> k;
  for (long long i = 0; i < n - 1; ++i) {
    long long a, b;
    std::cin >> a >> b;
    --a;
    --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  long long leaf = -1;
  for (long long i = 0; i < n; ++i) {
    if (g[i].size() == 2 || g[i].size() == 0) {
      std::cout << "NO";
      return 0;
    }
    if (g[i].size() == 1) {
      leaf = i;
    }
  }
  std::vector<long long> cur = bfs(leaf);
  long long maxI = std::max_element(cur.begin(), cur.end()) - cur.begin();
  if (cur[maxI] / 2 != k || cur[maxI] % 2 != 0) {
    std::cout << "NO";
    return 0;
  }
  dfs(leaf);
  long long dist = cur[maxI];
  while (dist > k) {
    dist--;
    maxI = pred[maxI];
  }
  std::vector<long long> cur2 = bfs(maxI);
  for (long long i = 0; i < n; ++i) {
    if (g[i].size() == 1 && cur2[i] != k) {
      std::cout << "NO";
      return 0;
    }
    if (g[i].size() != 1 && (i != maxI && (g[i].size() - 1) == 2)) {
      std::cout << "NO";
      return 0;
    }
  }
  std::cout << "YES";
}
