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;
const int N = 1e5 + 5;
int n, k, x, y, deg[N], root, u[N];
vector<vector<int> > g;
void dfs(int v, int dist, int p = 0) {
  if (dist == 0) {
    u[v] = 1;
    return;
  }
  for (auto to : g[v]) {
    if (to == p) continue;
    dfs(to, dist - 1, v);
    u[v] |= u[to];
  }
  if (u[v] && dist == k) root = v;
}
void Dfs(int v, int p = 0, int h = 0) {
  if (h > k || (deg[v] == 1 && h != k)) {
    puts("No");
    exit(0);
  }
  for (auto to : g[v]) {
    if (to == p) continue;
    Dfs(to, v, h + 1);
  }
}
int main() {
  cin >> n >> k;
  if (k > 10) {
    puts("No");
    return 0;
  }
  g.resize(n + 1);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    deg[x]++;
    deg[y]++;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      dfs(i, 2 * k);
      if (deg[root] < 3 || root == 0) {
        puts("No");
        return 0;
      }
      break;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i != root && deg[i] != 1 && deg[i] < 4) {
      puts("No");
      return 0;
    }
  }
  Dfs(root);
  puts("Yes");
}
