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 int maxn = 1e5 + 10;
vector<int> G[maxn];
int ru[maxn];
int cnt[maxn];
bool visit[maxn];
int main() {
  ios::sync_with_stdio(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
    ru[a]++;
    ru[b]++;
  }
  for (int i = 1; i <= k; i++) {
    memset(cnt, 0, sizeof cnt);
    for (int j = 1; j <= n; j++) {
      if (ru[j] == 1) {
        int len = G[j].size();
        int t;
        for (int l = 0; l <= len - 1; l++) {
          t = G[j][l];
          if (visit[t] != 1) {
            break;
          }
        }
        cnt[t]++;
        ru[j]--;
        visit[j] = true;
      }
    }
    int num = 0;
    for (int j = 1; j <= n; j++) ru[j] -= cnt[j];
    for (int j = 1; j <= n; j++) {
      if (cnt[j] != 0) {
        num++;
        if (cnt[j] < 3) {
          cout << "NO" << endl;
          return 0;
        }
      }
    }
    if (num < 3 && i != k) {
      cout << "No" << endl;
      return 0;
    }
    if (i == k) {
      if (num != 1) {
        cout << "No" << endl;
        return 0;
      }
    }
  }
  cout << "Yes" << endl;
}
