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 long long lmt = 1e5 + 5;
vector<long long> adj[lmt];
long long deg[lmt];
long long n, k, dis[lmt], root;
bool relax[lmt];
vector<long long> path;
void dfs(long long u, long long par) {
  for (long long v : adj[u]) {
    if (v == par) {
      continue;
    }
    dis[v] = dis[u] + 1;
    dfs(v, u);
  }
}
void get_path(long long u, long long des, long long par) {
  if (u == des) {
    relax[u] = 1;
    path.push_back(u);
  }
  for (long long v : adj[u]) {
    if (v == par) {
      continue;
    }
    get_path(v, des, u);
    if (relax[v]) {
      relax[u] = 1;
      path.push_back(u);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    deg[u]++, deg[v]++;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dis[1] = 0;
  dfs(1, -1);
  long long mx = 0, indx = 0, far = 0;
  for (long long i = 1; i <= n; i++) {
    if (dis[i] > mx) {
      mx = dis[i];
      indx = i;
    }
    dis[i] = 0;
  }
  dis[indx] = 0;
  dfs(indx, -1);
  mx = 0;
  for (long long i = 1; i <= n; i++) {
    if (dis[i] > mx) {
      mx = dis[i];
      far = i;
    }
    relax[i] = 0;
    dis[i] = 0;
  }
  get_path(indx, far, -1);
  root = path[path.size() / 2];
  dfs(root, -1);
  bool pos = 1;
  if (deg[root] < 3) {
    cout << "No"
         << "\n";
    return 0;
  }
  for (long long i = 1; i <= n; i++) {
    if (i == root) {
      continue;
    }
    if (deg[i] == 1) {
      if (dis[i] != k) {
        pos = 0;
        break;
      }
    } else {
      if (deg[i] < 4) {
        pos = 0;
        break;
      }
    }
  }
  if (pos) {
    cout << "Yes"
         << "\n";
  } else {
    cout << "No"
         << "\n";
  }
  return 0;
}
