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 MM = 120000;
vector<int> G[MM], T[MM];
int P[MM], L[MM];
void dfs(int u, int par) {
  P[u] = par;
  if (par != -1) L[u] = L[par] + 1;
  for (auto v : G[u]) {
    if (v != par) {
      dfs(v, u);
    }
  }
}
int N, K;
bool rip = 0;
void dfs2(int u, int par, int lev) {
  int kek = 0;
  for (auto v : G[u]) {
    if (v != par) {
      kek++;
      dfs2(v, u, lev + 1);
    }
  }
  if (kek < 3 && kek != 0) {
    rip = 1;
  }
  if (kek == 0 && lev != K) {
    rip = 1;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> K;
  for (int i = 0; i < N - 1; i++) {
    int u, v;
    cin >> u >> v;
    G[u - 1].push_back(v - 1);
    G[v - 1].push_back(u - 1);
  }
  dfs(0, -1);
  int mx = 0;
  int el = -1;
  for (int i = 0; i < N; i++) {
    mx = max(mx, L[i]);
    if (mx == L[i]) {
      el = i;
    }
  }
  if (mx >= K) {
    mx = K;
    while (mx--) el = P[el];
    int rt = el;
    dfs2(rt, -1, 0);
    if (rip) {
      cout << "No";
    } else {
      cout << "Yes";
    }
  } else {
    cout << "No";
    return 0;
  }
  return 0;
}
