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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
int const N = 1e5 + 20;
int n, k, h[N], par[N], mx[N];
vector<int> g[N];
void err() {
  cout << "No\n";
  exit(0);
}
void dfs1(int v) {
  for (int u : g[v]) {
    if (u != par[v]) {
      h[u] = h[v] + 1;
      par[u] = v;
      dfs1(u);
      mx[v] = max(mx[v], mx[u] + 1);
    }
  }
}
void dfs2(int v, int par = -1) {
  for (int u : g[v]) {
    if (u != par) {
      h[u] = h[v] + 1;
      dfs2(u, v);
    }
  }
}
void dfs3(int v, int par = -1) {
  if (par == -1) {
    if (g[v].size() < 3) err();
  } else {
    if (g[v].size() == 1) {
      if (h[v] != k) err();
    } else {
      if (g[v].size() < 4) err();
    }
  }
  for (int u : g[v])
    if (u != par) dfs3(u, v);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0, u, v; i < n - 1; i++) {
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  par[0] = -1;
  dfs1(0);
  int center = 0;
  while (mx[center] > k) {
    int v = center;
    for (int u : g[v]) {
      if (u != par[v] && mx[u] >= k) {
        center = u;
        break;
      }
    }
  }
  h[center] = 0;
  dfs2(center);
  dfs3(center);
  cout << "Yes\n";
}
