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;
mt19937 rng((int)std::chrono::steady_clock::now().time_since_epoch().count());
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  vector<vector<int> > g(n);
  vector<int> degree(n);
  for (int i = 0, u, v; i < n - 1; i++) {
    cin >> u >> v, u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
    degree[u]++;
    degree[v]++;
  }
  queue<int> bfs;
  for (int i = 0; i < n; i++) {
    if (degree[i] == 1) bfs.push(i);
  }
  vector<int> depth(n);
  vector<int> curdegree = degree;
  vector<bool> vis(n);
  while (!bfs.empty()) {
    int u = bfs.front();
    bfs.pop();
    degree[u] = 0;
    vis[u] = true;
    for (auto &v : g[u]) {
      if (vis[v]) continue;
      depth[v] = depth[u] + 1;
      degree[v]--;
      if (degree[v] == 1) {
        bfs.push(v);
      }
    }
  }
  map<int, vector<int> > depths;
  for (int i = 0; i < n; i++) {
    depths[depth[i]].push_back(i);
    if (depth[i] > k) {
      cout << "NO" << endl;
      return 0;
    } else if (depth[i] == k) {
      if (curdegree[i] < 3) {
        cout << "NO" << endl;
        return 0;
      }
    } else if (depth[i] > 0 and curdegree[i] < 4) {
      cout << "NO" << endl;
      return 0;
    }
    for (auto &v : g[i]) {
      if (abs(depth[i] - depth[v]) != 1) {
        cout << "NO" << endl;
        return 0;
      }
    }
  }
  for (int i = 0; i <= k; i++) {
    if (depths[i].size() == 0) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
}
