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> drogi[1000010];
int dist[2][1000010];
void dfs(int gdzie, int skad, int id) {
  for (int i : drogi[gdzie]) {
    if (i == skad) continue;
    dist[id][i] = dist[id][gdzie] + 1;
    dfs(i, gdzie, id);
  }
}
int k;
bool spr(int gdzie, int skad, int odl = 0) {
  if (odl > k) return 0;
  if (drogi[gdzie].size() == 1 && odl != k) return 0;
  for (int i : drogi[gdzie]) {
    if (skad == i) continue;
    if (!spr(i, gdzie, odl + 1)) return 0;
  }
  return 1;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n >> k;
  for (int a, b, i = 1; i < n; i++) {
    cin >> a >> b;
    drogi[a].push_back(b);
    drogi[b].push_back(a);
  }
  int ile = 0;
  int gdzie = 0;
  for (int i = 1; i <= n; i++) {
    if (drogi[i].size() == 1) {
      gdzie = i;
    }
    if (drogi[i].size() == 2) {
      cout << "No";
      return 0;
    }
    if (drogi[i].size() == 3) ile++;
  }
  if (ile > 1) {
    cout << "No";
    return 0;
  }
  dfs(gdzie, gdzie, 0);
  int pom = 1;
  for (int i = 1; i <= n; i++) {
    if (dist[0][i] == 2 * k) {
      pom = i;
    }
  }
  dfs(pom, pom, 1);
  for (int i = 1; i <= n; i++) {
    if (dist[0][i] == k && dist[1][i] == k) {
      if (ile && drogi[i].size() != 3) {
        cout << "No";
        return 0;
      }
      if (spr(i, i)) {
        cout << "Yes";
      } else {
        cout << "No";
      }
      return 0;
    }
  }
  cout << "No";
  return 0;
}
