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 comment(linker, "/STACK:1000000000")
using namespace std;
using pii = pair<int, int>;
using ll = long long;
using pll = pair<long long, long long>;
using pil = pair<int, long long>;
using vi = vector<int>;
using vll = vector<long long>;
using vs = vector<string>;
using vc = vector<char>;
using vb = vector<bool>;
using vpii = vector<pair<int, int>>;
using vd = vector<double>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
const ll mod = 1e9 + 7;
const ll INF = 1e18;
int n, k;
const int maxn = 1e5;
vi g[maxn];
int deg[maxn];
ll dist[maxn];
int par[maxn];
int bfs(int v) {
  memset(dist, 1e9, sizeof(dist));
  memset(par, -1, sizeof(par));
  dist[v] = 0;
  queue<int> q;
  q.push(v);
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    for (auto& to : g[cur]) {
      if (dist[to] or to == v) continue;
      dist[to] = dist[cur] + 1;
      par[to] = cur;
      q.push(to);
    }
  }
  return max_element(dist, dist + n) - dist;
}
void dfs(int v, int p = -1, int depth = 0) {
  if ((deg[v] == 1 and depth != k) or
      ((deg[v] - (p != -1)) < 3 and depth < k)) {
    cout << "No";
    exit(0);
  }
  for (auto& to : g[v]) {
    if (to == p) continue;
    dfs(to, v, depth + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[--u].push_back(--v);
    g[v].push_back(u);
    deg[u]++, deg[v]++;
  }
  int v1 = bfs(0);
  int v2 = bfs(v1);
  if (dist[v2] != 2LL * k) {
    cout << "No";
    return 0;
  }
  int cent = 0;
  int cur = v2;
  while (1) {
    if (dist[cur] == k) {
      cent = cur;
      break;
    }
    cur = par[cur];
    if (cur == -1) break;
  }
  dfs(cent);
  cout << "Yes";
  return 0;
}
