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 N = 1e5 + 5;
vector<int> g[N];
int c[N], p[N];
int main() {
  int n, k, u, v;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (g[i].size() == 1) q.push(i), c[i] = 1, p[i] = 2;
  int s = q.size(), l = -1, r;
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    s--;
    for (int i = 0; i < g[t].size(); i++) {
      if (p[g[t][i]] == 2) return cout << "No" && 0;
      if (!c[g[t][i]]) {
        q.push(g[t][i]);
        p[g[t][i]] = 1;
      }
      c[g[t][i]]++;
    }
    if (!s) {
      s = q.size();
      if (s) r = s;
      l++;
      for (int i = 1; i <= n; i++) {
        if (p[i] == 2) p[i] = 0;
        if (p[i] == 1 && c[i] < 3) return cout << "No" && 0;
        p[i] *= 2;
      }
    }
  }
  if (l == k && r == 1)
    cout << "Yes";
  else
    cout << "No";
}
