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>
#pragma GCC optimize("O3")
using namespace std;
int n, k, gr[100002];
vector<int> v[100002];
int fii[100002];
int hdg[100002], pusi;
int pus[100002];
bool tt(int nod) {
  for (int i = 0; i < v[nod].size(); ++i)
    if (!pus[v[nod][i]]) return 1;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  cin >> n >> k;
  if (n <= 3) {
    cout << "No";
    return 0;
  }
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
    gr[a]++;
    gr[b]++;
  }
  queue<int> d;
  for (int i = 1; i <= n; ++i)
    if (gr[i] == 1) d.push(i), ++pusi, pus[i] = 1;
  while (!d.empty()) {
    int x = d.front();
    if (hdg[x] >= 1 && fii[x] < 3) {
      cout << "No";
      return 0;
    }
    if (hdg[x] > k) {
      cout << "No";
      return 0;
    }
    if (pusi == n && hdg[x] != k) {
      cout << "No";
      return 0;
    }
    d.pop();
    for (int i = 0; i < v[x].size(); ++i) {
      int vecin = v[x][i];
      if (gr[vecin]) {
        ++fii[vecin];
        --gr[x];
        --gr[vecin];
        int vl = hdg[x] + 1;
        if (hdg[vecin] == 0)
          hdg[vecin] = vl;
        else if (vl != hdg[vecin]) {
          cout << "No";
          return 0;
        }
        if (gr[vecin] == 1) {
          if (tt(vecin)) d.push(vecin), ++pusi, pus[vecin] = 1;
        } else if (gr[vecin] == 0)
          d.push(vecin), ++pusi, pus[vecin] = 1;
      }
    }
  }
  cout << "Yes";
  return 0;
}
