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>
const int N = (int)1e5 + 7;
const int inf = (int)1e9 + 7;
const int mod = (int)1e9 + 7;
const long long linf = (long long)1e18 + 7;
const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
using namespace std;
int n, k;
vector<int> g[N];
int deg[N], d[N];
void findCenter() {
  fill(d + 1, d + 1 + n, -1);
  queue<int> q;
  for (int i = (1); i <= (n); i++) {
    if ((int)g[i].size() == 1) {
      d[i] = 0;
      q.push(i);
    }
  }
  int root = -1;
  while ((int)q.size()) {
    int v = q.front();
    q.pop();
    root = v;
    if (d[v] > k || (d[v] > 0 && deg[v] < 3)) cout << "No", exit(0);
    int ch = 0;
    for (auto to : g[v]) {
      if (d[to] == d[v] - 1 && d[v] > 0) continue;
      if (d[to] == -1) {
        d[to] = d[v] + 1;
        q.push(to);
        deg[to]++;
        ++ch;
      } else if (d[to] == d[v] + 1) {
        deg[to]++;
        ++ch;
      } else {
        cout << "No", exit(0);
      }
    }
    if (ch > 1) cout << "No", exit(0);
  }
  if (d[root] != k) cout << "No", exit(0);
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = (1); i <= (n - 1); i++) {
    int v, u;
    cin >> v >> u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  findCenter();
  cout << "Yes";
  exit(0);
}
