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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const int N = 1e5 + 7;
vector<int> g[N];
int sz[N], lvl[N];
int cur_sz[N];
bool used[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  if (n < 3) {
    cout << "No\n";
    return 0;
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  vector<int> leaf(n, false);
  vector<int> cur;
  for (int i = 0; i < n; i++) {
    sz[i] = g[i].size();
    cur_sz[i] = g[i].size();
    if (g[i].size() == 1) {
      used[i] = true;
      leaf[i] = true;
      cur.push_back(i);
    }
  }
  while (true) {
    set<int> kek;
    for (int v : cur) {
      for (int to : g[v]) {
        if (used[to]) continue;
        if (lvl[to] != 0 && lvl[to] != lvl[v] + 1) {
          cout << "No\n";
          return 0;
        }
        lvl[to] = lvl[v] + 1;
        cur_sz[to]--;
        if (cur_sz[to] == 1) {
          kek.insert(to);
        }
        if (cur_sz[to] != 1 && kek.count(to)) {
          kek.erase(to);
        }
      }
      cur_sz[v] = 0;
    }
    cur.clear();
    for (int x : kek) {
      used[x] = true;
      cur.push_back(x);
    }
    if (cur.empty()) break;
  }
  int mx = -1;
  int ind = 0;
  for (int i = 0; i < n; i++) {
    if (lvl[i] > mx) {
      ind = i;
    }
    mx = max(mx, lvl[i]);
  }
  if (mx != k) {
    cout << "No\n";
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (i != ind && lvl[i] == mx) {
      cout << "No\n";
      return 0;
    }
    if (leaf[i]) continue;
    int need = 4;
    need -= (ind == i);
    if (sz[i] < need) {
      cout << "No\n";
      return 0;
    }
  }
  cout << "Yes\n";
  return 0;
}
