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 diam, diamcv, center;
vector<int> g[100005];
void dfs(int x, int p, int d) {
  if (d > diam) {
    diam = d;
    diamcv = x;
  }
  for (int i = 0; i < g[x].size(); i++)
    if (g[x][i] != p) dfs(g[x][i], x, d + 1);
}
int dfs2(int x, int p) {
  if (x == diamcv) return -diam / 2;
  for (int i = 0; i < g[x].size(); i++)
    if (g[x][i] != p) {
      int res = dfs2(g[x][i], x);
      if (res == 0) return 0;
      if (res == -1) {
        center = x;
        return 0;
      }
      if (res != -1e9 && res < 0) return res + 1;
    }
  return -1e9;
}
bool dfs3(int c, int p, int k) {
  if (k == 0) return g[c].size() == 1;
  if (g[c].size() < 4) return false;
  bool res = true;
  for (int i = 0; i < g[c].size(); i++)
    if (g[c][i] != p) res &= dfs3(g[c][i], c, k - 1);
  return res;
}
int n, k, u, v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int first;
  for (int i = 1; i <= n; i++)
    if (g[i].size() == 1) {
      first = i;
      dfs(i, 0, 0);
      break;
    }
  if (diam % 2) {
    cout << "No";
    return 0;
  }
  dfs2(first, 0);
  if (g[center].size() < 3) {
    cout << "No";
    return 0;
  }
  bool res = true;
  for (int i = 0; i < g[center].size(); i++)
    res &= dfs3(g[center][i], center, k - 1);
  cout << (res ? "Yes" : "No");
  return 0;
}
