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;
inline int gi() {
  int data = 0, fu = 1;
  char ch = 0;
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') fu = -1, ch = getchar();
  while ('0' <= ch && ch <= '9')
    data = (data << 1) + (data << 3) + (ch ^ 48), ch = getchar();
  return data * fu;
}
int n, K, deg[100005];
vector<int> edge[100005];
queue<int> Q;
void dfs(int u, int From, int deep) {
  if (edge[u].size() == 1) {
    if (deep != K) puts("No"), exit(0);
    return;
  }
  int c = 0;
  for (auto v : edge[u])
    if (v != From) dfs(v, u, deep + 1), ++c;
  if (c < 3) return puts("No"), exit(0);
}
int main() {
  n = gi();
  K = gi();
  for (int i = 1, u, v; i < n; ++i) {
    u = gi();
    v = gi();
    ++deg[u];
    ++deg[v];
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i)
    if (deg[i] != 1 && deg[i] < 3) return puts("No"), 0;
  for (int i = 1; i <= n; ++i)
    if (deg[i] == 1) Q.push(i);
  int root = 0;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    root = u;
    for (auto v : edge[u]) {
      --deg[v];
      if (deg[v] == 1) Q.push(v);
    }
  }
  dfs(root, 0, 0);
  puts("Yes");
  return 0;
}
