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 maxn = 1e5 + 9;
int N, K, deg[maxn], d[maxn], cnt[maxn], vis[maxn];
vector<int> g[maxn];
queue<int> Q;
void No() { exit(0 * puts("No")); }
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1, u, v; i < N; i++) {
    scanf("%d%d", &u, &v);
    deg[--u]++;
    deg[--v]++;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < N; i++)
    if (deg[i] == 1) Q.push(i);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    vis[x] = 1;
    if (d[x] > 0 && cnt[x] < 3) No();
    for (auto e : g[x]) {
      if (vis[e]) continue;
      if (d[e] == 0)
        d[e] = d[x] + 1;
      else if (d[e] != d[x] + 1)
        No();
      cnt[e]++;
      if (--deg[e] == 1) Q.push(e);
    }
  }
  puts(*max_element(d, d + N) == K ? "Yes" : "No");
  return 0;
}
