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 = 100 * 1000 + 10;
int deg[N];
vector<int> G[N];
int center = -1;
vector<int> verts;
void find_center(int x, int k, int pre) {
  verts.push_back(x);
  if ((int)verts.size() == 2 * k + 1) {
    center = verts[k];
    return;
  }
  for (int y : G[x]) {
    if (y != pre) find_center(y, k, x);
  }
  verts.pop_back();
}
bool is_hedge(int x, int k, int pre) {
  if (k == 0) {
    return G[x].size() == 1 && G[x][0] == pre;
  }
  bool all = true;
  for (int y : G[x]) {
    if (y != pre) all &= is_hedge(y, k - 1, x);
  }
  return all && ((pre == -1 && G[x].size() >= 3) || G[x].size() > 3);
}
int main() {
  do {
    ios::sync_with_stdio(0);
    cin.tie(0);
  } while (0);
  int n;
  int k;
  cin >> n >> k;
  for (int i = 0; i < (n - 1); i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      find_center(i, k, -1);
      break;
    }
  }
  if (center != -1 && is_hedge(center, k, -1))
    cout << "Yes\n";
  else
    cout << "No\n";
}
