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;
vector<int> v[100001];
bool visit[100001];
int dist[100001];
int dig[100001];
int child[100001];
void dfs(int x) {
  visit[x] = true;
  for (int i = 0; i < v[x].size(); i++) {
    if (!visit[v[x][i]]) {
      dist[v[x][i]] = dist[x] + 1;
      dfs(v[x][i]);
      child[x] += child[v[x][i]];
    }
  }
  child[x]++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
    dig[x]++;
    dig[y]++;
  }
  for (int i = 1; i <= n; i++) {
    if (dig[i] == 1) {
      dfs(i);
      break;
    }
  }
  int center = -1, mx = 0;
  for (int i = 1; i <= n; i++) {
    if (dist[i] == k && dig[i] >= 3) {
      if (child[i] > mx) {
        mx = child[i];
        center = i;
      }
    }
  }
  if (center == -1) {
    cout << "NO";
    return 0;
  }
  memset(dist, 0, sizeof(dist));
  memset(visit, false, sizeof(visit));
  dfs(center);
  for (int i = 1; i <= n; i++) {
    if (dist[i] < k && i != center) {
      if (dig[i] < 4) {
        cout << "NO";
        return 0;
      }
    } else if (dist[i] == k && i != center) {
      if (dig[i] != 1) {
        cout << "NO";
        return 0;
      }
    } else if (dist[i] > k) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES";
}
