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;
long long int cons;
inline long long int check(long long int a) {
  if (a >= cons) a %= cons;
  return a;
}
inline long long int check2(long long int a) {
  a %= cons;
  if (a < 0) a += cons;
  return a;
}
inline long long int fast_check(long long int a) {
  if (a >= cons) a -= cons;
  return a;
}
long long int GCD(long long int a, long long int b) {
  while (a && b) {
    a = a % b;
    if (a != 0) b = b % a;
  }
  return a + b;
}
long long int exp(long long int a, long long int n) {
  long long int ans = 1;
  a = check(a);
  while (n) {
    if (n & 1) ans = check(ans * a);
    a = check(a * a);
    n = (n >> 1);
  }
  return ans;
}
void print_pair(pair<int, int> p) {
  cout << p.first << " " << p.second << endl;
}
long long int nc2(long long int a) { return (a * (a - 1)) / 2; }
long long int nc3(long long int a) {
  long long int b = a * (a - 1) * (a - 2);
  return b / 6;
}
const int N = 1e5 + 10;
vector<int> graph[N];
bool visited[N];
int deg[N];
bool processed[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cons = 1e9 + 7;
  srand(time(NULL));
  int n, k;
  cin >> n >> k;
  int coun = 0;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  vector<int> v1, v2;
  for (int i = 1; i <= n; i++) {
    if (graph[i].size() == 1) {
      coun++;
      v1.push_back(i);
      visited[i] = true;
    }
  }
  int tim = 0;
  while (v1.size() > 1) {
    vector<int> temp;
    temp.resize(v1.size());
    int first = -1;
    tim++;
    for (auto itr : v1) {
      first++;
      for (auto nei : graph[itr]) {
        if (visited[nei]) continue;
        temp[first]++;
        if (!processed[nei]) {
          coun++;
          v2.push_back(nei);
        }
        processed[nei] = true;
        deg[nei]++;
      }
    }
    for (auto itr : temp) {
      if (itr != 1) {
        cout << "No";
        return 0;
      }
    }
    for (auto itr : v2) {
      if (deg[itr] < 3) {
        cout << "No";
        return 0;
      }
      visited[itr] = true;
    }
    v1 = v2;
    v2.clear();
  }
  if (tim == k && coun == n && v1.size() == 1) {
    cout << "Yes";
  } else
    cout << "No";
  return 0;
}
