#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 7;
const int MX = 1e9 + 7;
const long long int INF = 1e18 + 9LL;
int n, k;
int d[N];
int cnt[N];
queue<int> Q;
vector<int> G[N];
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i)
    if (G[i].size() == 1)
      Q.push(i);
    else
      d[i] = -1;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int v : G[u]) {
      if (d[v] == -1) {
        d[v] = d[u] + 1;
        Q.push(v);
      }
      if (d[v] == d[u] + 1) cnt[v]++;
    }
  }
  int cntk = 0;
  for (int i = 1; i <= n; ++i) {
    if (d[i] == k) ++cntk;
    if (d[i] > k) {
      puts("No");
      return 0;
    }
    if (d[i] != 0 && cnt[i] < 3) {
      puts("No");
      return 0;
    }
    for (int v : G[i])
      if (d[v] == d[i]) {
        puts("No");
        return 0;
      }
  }
  if (cntk == 1)
    puts("Yes");
  else
    puts("No");
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
const int N = 1e6 + 7;
const int MX = 1e9 + 7;
const long long int INF = 1e18 + 9LL;
int n, k;
int d[N];
int cnt[N];
queue<int> Q;
vector<int> G[N];
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i)
    if (G[i].size() == 1)
      Q.push(i);
    else
      d[i] = -1;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int v : G[u]) {
      if (d[v] == -1) {
        d[v] = d[u] + 1;
        Q.push(v);
      }
      if (d[v] == d[u] + 1) cnt[v]++;
    }
  }
  int cntk = 0;
  for (int i = 1; i <= n; ++i) {
    if (d[i] == k) ++cntk;
    if (d[i] > k) {
      puts("No");
      return 0;
    }
    if (d[i] != 0 && cnt[i] < 3) {
      puts("No");
      return 0;
    }
    for (int v : G[i])
      if (d[v] == d[i]) {
        puts("No");
        return 0;
      }
  }
  if (cntk == 1)
    puts("Yes");
  else
    puts("No");
  return 0;
}

```