#include <bits/stdc++.h>
using namespace std;
set<int> second[100005];
bool used[100005];
int c[100005];
vector<int> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    second[x].insert(y);
    second[y].insert(x);
  }
  for (int i = 1; i <= n; i++) {
    if (second[i].size() == 1) {
      v.push_back(i);
      used[i] = true;
    }
  }
  while (k >= 1) {
    if (v.empty()) {
      cout << "No" << endl;
      return 0;
    }
    vector<int> nv;
    for (int x : v) {
      int y = *second[x].begin();
      c[y]++;
      second[x].erase(y);
      second[y].erase(x);
      if (used[y] == false) {
        used[y] = true;
        nv.push_back(y);
      }
    }
    for (int x : nv) {
      if (c[x] < 3 || second[x].size() != min(1, k - 1)) {
        cout << "No" << endl;
        return 0;
      }
    }
    v = nv;
    k--;
  }
  cout << "Yes" << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
set<int> second[100005];
bool used[100005];
int c[100005];
vector<int> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    second[x].insert(y);
    second[y].insert(x);
  }
  for (int i = 1; i <= n; i++) {
    if (second[i].size() == 1) {
      v.push_back(i);
      used[i] = true;
    }
  }
  while (k >= 1) {
    if (v.empty()) {
      cout << "No" << endl;
      return 0;
    }
    vector<int> nv;
    for (int x : v) {
      int y = *second[x].begin();
      c[y]++;
      second[x].erase(y);
      second[y].erase(x);
      if (used[y] == false) {
        used[y] = true;
        nv.push_back(y);
      }
    }
    for (int x : nv) {
      if (c[x] < 3 || second[x].size() != min(1, k - 1)) {
        cout << "No" << endl;
        return 0;
      }
    }
    v = nv;
    k--;
  }
  cout << "Yes" << endl;
  return 0;
}

```