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 N = 1e5 + 10;
int power(int a, int b, int mod) {
  int res = 1;
  while (b) {
    if (b & 1) {
      res = 1LL * res * a % mod;
    }
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return res;
}
int n, k, root = -1, fail = 0;
vector<int> adj[N];
int dfs(int node = 1, int par = 0) {
  int max_len = 0;
  for (int child : adj[node]) {
    if (child == par) continue;
    max_len = max(max_len, dfs(child, node));
  }
  if (max_len == k) {
    root = node;
  }
  return max_len + 1;
}
int dfs1(int node, int par, int h) {
  int val = (node == root ? 3 : 4);
  if (h < 0 || (h > 0 && adj[node].size() < val)) fail = 1;
  for (int child : adj[node]) {
    if (child == par) continue;
    fail |= dfs1(child, node, h - 1);
  }
  return fail;
}
int main() {
  int u, v;
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs();
  if (root == -1) {
    printf("No\n");
    return 0;
  }
  fail |= dfs1(root, 0, k);
  if (fail)
    printf("No\n");
  else {
    printf("Yes\n");
  }
  return 0;
}
