#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7;
int n, k, u, v, flag, du[maxn];
set<int> head[maxn];
map<int, int> M;
void add() {
  head[u].insert(v);
  head[v].insert(u);
}
int sovle() {
  for (int i = 1; i <= n; i++)
    if (head[i].size() == 1) M[i] = 1;
  while (n > 1) {
    map<int, int> Tm;
    for (auto i : M) {
      int x = *head[i.first].begin();
      Tm[x]++;
      n--;
      head[x].erase(i.first);
    }
    for (auto i : Tm)
      if (i.second < 3) return 0 * printf("No\n");
    M = Tm;
    k--;
  }
  return 0 * printf("%s\n", k ? "No" : "Yes");
}
void input() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) cin >> u >> v, add();
}
int main() {
  input();
  sovle();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7;
int n, k, u, v, flag, du[maxn];
set<int> head[maxn];
map<int, int> M;
void add() {
  head[u].insert(v);
  head[v].insert(u);
}
int sovle() {
  for (int i = 1; i <= n; i++)
    if (head[i].size() == 1) M[i] = 1;
  while (n > 1) {
    map<int, int> Tm;
    for (auto i : M) {
      int x = *head[i.first].begin();
      Tm[x]++;
      n--;
      head[x].erase(i.first);
    }
    for (auto i : Tm)
      if (i.second < 3) return 0 * printf("No\n");
    M = Tm;
    k--;
  }
  return 0 * printf("%s\n", k ? "No" : "Yes");
}
void input() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) cin >> u >> v, add();
}
int main() {
  input();
  sovle();
  return 0;
}

```