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 MAXN = 1e5 + 10;
int n;
vector<int> g[MAXN];
int distX[MAXN], distY[MAXN];
void dfs(int v, int par, int* dist) {
  dist[v] = dist[par] + (par == 0 ? 0 : 1);
  for (int w : g[v])
    if (w != par) dfs(w, v, dist);
}
int check(int v, int k, int par) {
  if (k == 0) return 1;
  if (g[v].size() <= 3) {
    return 0;
  }
  int res = 1;
  for (int w : g[v]) {
    if (w == par) continue;
    res += check(w, k - 1, v);
  }
  return res;
}
int main() {
  int k;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int x = 1;
  for (int i = 1; i <= n; i++)
    if (g[i].size() == 1) x = i;
  dfs(x, 0, distX);
  int y = 0, d = 0;
  for (int i = 1; i <= n; i++) {
    if (d < distX[i]) {
      d = distX[i];
      y = i;
    }
  }
  dfs(y, 0, distY);
  int z = 1;
  for (int i = 1; i <= n; i++)
    if (distX[i] == d / 2 && distY[i] == d / 2) z = i;
  g[z].push_back(0);
  int ans = check(z, k, 0);
  if (ans == n)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}
