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>
const double eps = (1e-9);
using namespace std;
int dcmp(long double a, long double b) {
  return fabsl(a - b) <= eps ? 0 : (a > b) ? 1 : -1;
}
int getBit(long long num, int idx) { return ((num >> idx) & 1ll) == 1; }
int setBit1(int num, int idx) { return num | (1 << idx); }
long long setBit0(long long num, int idx) { return num & ~(1ll << idx); }
long long flipBit(long long num, int idx) { return num ^ (1ll << idx); }
void M() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int countNumBit1(int mask) {
  int ret = 0;
  while (mask) {
    mask &= (mask - 1);
    ++ret;
  }
  return ret;
}
int n, m, mxl, Node, vis[100009];
vector<vector<int> > v(100009);
vector<int> v1, myv;
void dfs1(int no, int lv) {
  vis[no] = 1;
  if (lv > mxl) {
    mxl = lv;
    Node = no;
  }
  for (int i = 0; i < ((int)(v[no]).size()); i++) {
    int nw = v[no][i];
    if (vis[nw] == 1) continue;
    dfs1(nw, lv + 1);
  }
}
void dfs2(int S, int E) {
  vis[S] = 1;
  if (S == E) {
    myv = v1;
    return;
  }
  for (int i = 0; i < ((int)(v[S]).size()); i++) {
    int nw = v[S][i];
    if (vis[nw]) continue;
    v1.push_back(nw);
    dfs2(nw, E);
    v1.pop_back();
  }
}
void bfs(int root) {
  queue<int> q;
  q.push(root);
  vis[root] = 1;
  while (!q.empty()) {
    int siz = ((int)(q).size());
    bool fl = 0, one = 0;
    while (siz--) {
      int cur = q.front();
      q.pop();
      bool sub = (cur != root);
      if (((int)(v[cur]).size()) - sub < 3) fl = 1;
      if (((int)(v[cur]).size()) - sub > 0) one = 1;
      for (int i = 0; i < ((int)(v[cur]).size()); i++) {
        int nw = v[cur][i];
        if (vis[nw]) continue;
        vis[nw] = 1;
        q.push(nw);
      }
    }
    if (fl && one) {
      cout << "NO\n";
      return;
    }
  }
  cout << "YES\n";
}
int main() {
  int k;
  cin >> n >> k;
  int f, t;
  for (int i = 0; i < n - 1; i++) {
    cin >> f >> t;
    v[f].push_back(t);
    v[t].push_back(f);
  }
  mxl = 0;
  dfs1(1, 1);
  memset(vis, 0, sizeof(vis));
  int n1 = Node;
  mxl = 0;
  dfs1(n1, 1);
  int n2 = Node;
  memset(vis, 0, sizeof(vis));
  v1.push_back(n1);
  dfs2(n1, n2);
  v1.pop_back();
  int mid = ((int)(myv).size()) / 2;
  if (((int)(myv).size()) % 2 == 0 || mid != k) {
    cout << "NO\n";
    return 0;
  }
  int root = myv[mid];
  memset(vis, 0, sizeof(vis));
  bfs(root);
}
