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 long long inff = 0x3f3f3f3f3f3f3f3f;
int n, k, zz = 0, du[100008], x, y, b[200008], sd, vis[100008], c[100008],
          vv[100008];
set<pair<int, int> > se;
vector<int> g[100008];
int main() {
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  if (n < 4) return puts("No"), 0;
  for (int i(1); i <= (n - 1); ++i) {
    scanf("%d", &x), scanf("%d", &y);
    g[x].push_back(y);
    g[y].push_back(x);
    du[x]++, du[y]++;
  }
  for (int i(1); i <= (n); ++i) se.insert(pair<int, int>(du[i], i));
  while (!se.empty()) {
    pair<int, int> qw = *se.begin();
    if (qw.first == 1) {
      for (int j = 0; j < g[qw.second].size(); j++) {
        int v = g[qw.second][j];
        if (vis[v]) continue;
        b[++sd] = v;
        c[v]++;
      }
      vis[qw.second] = 1;
      se.erase(se.begin());
    } else {
      for (int i(1); i <= (sd); ++i)
        if (c[b[i]] < 3) return puts("No"), 0;
      for (int i(1); i <= (sd); ++i) {
        if (!vv[b[i]]) {
          vv[b[i]] = 1;
          se.erase(se.find(pair<int, int>(du[b[i]], b[i])));
          du[b[i]] -= c[b[i]];
          se.insert(pair<int, int>(du[b[i]], b[i]));
        }
      }
      zz++;
      sd = 0;
      if (se.size() == 1) break;
    }
  }
  if (zz == k && se.size() == 1)
    puts("Yes");
  else
    puts("No");
  return 0;
}
