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;
long long int Pow(long long int a, long long int b, long long int md,
                  long long int ans = 1) {
  for (; b; b >>= 1, a = a * a % md)
    if (b & 1) ans = ans * a % md;
  return ans % md;
}
const long long int MAXN = 1e6 + 10;
const long long int INF = 8e18;
const long long int MOD = 1e9 + 7;
long long int cnt[MAXN], deg[MAXN], dist[MAXN], n, k;
set<long long int> adj[MAXN];
void BFS() {
  fill(dist, dist + MAXN, INF);
  queue<long long int> Q;
  for (long long int i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      Q.push(i);
      dist[i] = 0;
    }
  }
  while ((long long int)Q.size()) {
    long long int v = Q.front();
    if (deg[v] == 0) break;
    long long int u = *adj[v].begin();
    Q.pop();
    cnt[u]++;
    if (dist[u] != dist[v] + 1 && dist[u] != INF) {
      printf("No\n");
      exit(0);
    }
    dist[u] = dist[v] + 1;
    adj[u].erase(v), deg[u]--;
    if (deg[u] == 1) Q.push(u);
  }
}
int main() {
  scanf("%lld%lld", &n, &k);
  for (long long int i = 1; i <= n - 1; i++) {
    long long int u, v;
    scanf("%lld%lld", &u, &v);
    adj[u].insert(v);
    adj[v].insert(u);
    deg[u]++, deg[v]++;
  }
  BFS();
  long long int mx = 0;
  for (long long int i = 1; i <= n; i++)
    if (dist[i] != INF) mx = max(mx, dist[i]);
  for (long long int i = 1; i <= n; i++) {
    if (cnt[i] > 0 && cnt[i] < 3) return !printf("No\n");
  }
  printf(mx == k ? "Yes\n" : "No\n");
  return 0;
}
