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 c = 100002;
int n, k, dist[c][3];
bool v[c][3], jo, baj;
vector<int> sz[c];
void dfs(int a, int b) {
  v[a][b] = true;
  for (int i = 0; i < sz[a].size(); i++) {
    int x = sz[a][i];
    if (!v[x][b]) dist[x][b] = dist[a][b] + 1, dfs(x, b);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    sz[x].push_back(y), sz[y].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (sz[i].size() == 1) {
      dfs(i, 0);
      break;
    }
  for (int i = 1; i <= n; i++)
    if (dist[i][0] == 2 * k) {
      jo = 1, dfs(i, 1);
      break;
    }
  for (int i = 1; i <= n; i++)
    if (dist[i][0] == k && dist[i][1] == k) dfs(i, 2);
  for (int i = 1; i <= n; i++) {
    int x = dist[i][2], y = sz[i].size();
    if ((!x && y < 3) || (0 < x && x < k && y < 4) || (x == k && y != 1) ||
        x > k)
      baj = 1;
  }
  if (jo && !baj)
    cout << "Yes\n";
  else
    cout << "No\n";
  return 0;
}
