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;
vector<int> v[100009];
int n, cc, cd, pr[100009];
void dfsc(int x, int par, int d) {
  pr[x] = par;
  if (d > cd) {
    cd = d;
    cc = x;
  }
  for (int i = 0; i < v[x].size(); i++) {
    int y = v[x][i];
    if (y == par) continue;
    dfsc(y, x, d + 1);
  }
}
int findcenter() {
  dfsc(0, 0, 0);
  cd = 0;
  dfsc(cc, cc, 0);
  for (int i = 0; i < cd / 2; i++) {
    cc = pr[cc];
  }
  return cc;
}
int dep[100009];
void no() {
  cout << "NO";
  exit(0);
}
int k;
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int x, y, i = 1; i < n; i++) {
    cin >> x >> y;
    x--;
    y--;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  int c = findcenter();
  queue<int> q;
  q.push(c);
  pr[c] = c;
  dep[c] = 0;
  int forb = -1;
  while (!q.empty()) {
    int x = q.front(), par = pr[x];
    q.pop();
    int s = 0;
    for (int i = 0; i < v[x].size(); i++) {
      int y = v[x][i];
      if (y == par) continue;
      pr[y] = x;
      dep[y] = dep[x] + 1;
      q.push(y);
      s++;
    }
    if (s >= 3) continue;
    if (s == 0) {
      if (forb == -1)
        forb = dep[x];
      else if (forb != dep[x])
        no();
    } else
      no();
  }
  if (k != forb) no();
  cout << "YES";
}
