#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353ll;
const long long N = 100050;
long long n, k;
vector<long long> g[N];
stack<long long> path;
long long beginn, endd;
long long maxx = 0;
void diameter(long long node, long long par, long long dist) {
  for (auto a : g[node]) {
    if (a != par) diameter(a, node, dist + 1);
  }
  if (dist >= maxx) {
    maxx = dist;
    beginn = node;
  }
}
void diameter1(long long node, long long par, long long dist) {
  for (auto a : g[node]) {
    if (a != par) diameter1(a, node, dist + 1);
  }
  if (dist >= maxx) {
    maxx = dist;
    endd = node;
  }
}
bool find_path(long long node, long long par, long long dest) {
  path.push(node);
  if (node == dest) return 1;
  for (auto a : g[node]) {
    if (a == par) continue;
    if (find_path(a, node, dest)) return 1;
  }
  path.pop();
  return 0;
}
long long find_root() {
  diameter(1, -1, 0);
  diameter1(beginn, -1, 0);
  find_path(beginn, -1, endd);
  long long kk = k;
  while (kk-- && !path.empty()) {
    path.pop();
  }
  return path.empty() ? -1 : path.top();
}
bool check(long long node, long long par, long long d) {
  if (par == -1) {
    if (g[node].size() < 3) return 0;
  } else {
    if (g[node].size() < 4 && g[node].size() != 1) return 0;
  }
  if (g[node].size() == 1) {
    if (d != k) return 0;
  }
  for (auto a : g[node]) {
    if (a == par) continue;
    if (!check(a, node, d + 1)) return 0;
  }
  return 1;
}
signed main() {
  ios::sync_with_stdio(0);
  cin >> n >> k;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  long long root = find_root();
  if (root != -1 && check(root, -1, 0))
    cout << "Yes";
  else
    cout << "No";
  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 long long mod = 998244353ll;
const long long N = 100050;
long long n, k;
vector<long long> g[N];
stack<long long> path;
long long beginn, endd;
long long maxx = 0;
void diameter(long long node, long long par, long long dist) {
  for (auto a : g[node]) {
    if (a != par) diameter(a, node, dist + 1);
  }
  if (dist >= maxx) {
    maxx = dist;
    beginn = node;
  }
}
void diameter1(long long node, long long par, long long dist) {
  for (auto a : g[node]) {
    if (a != par) diameter1(a, node, dist + 1);
  }
  if (dist >= maxx) {
    maxx = dist;
    endd = node;
  }
}
bool find_path(long long node, long long par, long long dest) {
  path.push(node);
  if (node == dest) return 1;
  for (auto a : g[node]) {
    if (a == par) continue;
    if (find_path(a, node, dest)) return 1;
  }
  path.pop();
  return 0;
}
long long find_root() {
  diameter(1, -1, 0);
  diameter1(beginn, -1, 0);
  find_path(beginn, -1, endd);
  long long kk = k;
  while (kk-- && !path.empty()) {
    path.pop();
  }
  return path.empty() ? -1 : path.top();
}
bool check(long long node, long long par, long long d) {
  if (par == -1) {
    if (g[node].size() < 3) return 0;
  } else {
    if (g[node].size() < 4 && g[node].size() != 1) return 0;
  }
  if (g[node].size() == 1) {
    if (d != k) return 0;
  }
  for (auto a : g[node]) {
    if (a == par) continue;
    if (!check(a, node, d + 1)) return 0;
  }
  return 1;
}
signed main() {
  ios::sync_with_stdio(0);
  cin >> n >> k;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  long long root = find_root();
  if (root != -1 && check(root, -1, 0))
    cout << "Yes";
  else
    cout << "No";
  return 0;
}

```