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;
int n, k;
vector<vector<int>> A;
vector<int> D;
vector<bool> V;
int f(int v) {
  V[v] = true;
  if (A[v].size() == 1) return D[v] = 0;
  int d = 1 << 30;
  for (auto& u : A[v])
    if (!V[u]) d = min(d, f(u) + 1);
  return D[v] = d;
}
int main() {
  cin >> n >> k;
  A.resize(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    A[a].push_back(b);
    A[b].push_back(a);
  }
  D.assign(n + 1, -1);
  for (int i = 1; i <= n; i++) {
    if (D[i] == -1) {
      V.assign(n + 1, false);
      D[i] = f(i);
    }
  }
  int m = 0;
  for (int i = 1; i <= n; i++) {
    if (D[m] < D[i]) {
      m = i;
    }
  }
  if (D[m] != k) {
    cout << "No" << '\n';
    return 0;
  }
  queue<int> Q;
  Q.push(m);
  while (!Q.empty()) {
    int v = Q.front();
    Q.pop();
    if (A[v].size() != 1) {
      if (A[v].size() < 4 and v != m) {
        cout << "No" << '\n';
        return 0;
      }
      if (A[v].size() < 3 and v == m) {
        cout << "No" << '\n';
        return 0;
      }
    }
    bool p = false;
    for (auto a : A[v]) {
      if (D[a] == D[v] - 1) {
        Q.push(a);
        continue;
      }
      if (D[a] == D[v] + 1 and !p) {
        p = true;
        continue;
      }
      cout << "No" << '\n';
      return 0;
    }
  }
  cout << "Yes" << '\n';
}
