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>
#pragma GCC optimize "-O3"
using namespace std;
const int MAXN = 120000;
int n, k;
int h[MAXN];
vector<int> eds[MAXN];
int pp[MAXN];
void dfs1(int v, int p) {
  for (int u : eds[v]) {
    if (u == p) continue;
    h[u] = h[v] + 1;
    pp[u] = v;
    dfs1(u, v);
  }
}
void no() {
  cout << "No\n";
  exit(0);
}
void dfs2(int v, int p) {
  int cc = 0;
  for (int u : eds[v]) {
    if (u == p) continue;
    ++cc;
    h[u] = h[v] + 1;
    dfs2(u, v);
  }
  if (cc == 0) {
    if (h[v] != k) no();
  } else if (cc < 3) {
    no();
  }
}
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) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    eds[a].push_back(b);
    eds[b].push_back(a);
  }
  dfs1(0, -1);
  int mx = -1;
  for (int i = 0; i < n; ++i)
    if (mx == -1 || h[i] > h[mx]) mx = i;
  if (h[mx] < k) no();
  for (int i = 0; i < k; ++i) mx = pp[mx];
  h[mx] = 0;
  dfs2(mx, -1);
  cout << "Yes\n";
  return 0;
}
