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 + 1;
int n, m, dist[N], root = -1;
vector<int> ve[N];
bool ok = 1, ok1 = 0;
stack<int> st;
void dfs(int u, int par) {
  if (ok1) return;
  if (st.size() == 2 * m - 1) {
    while (st.size() > m) st.pop();
    root = st.top();
    ok1 = 1;
    return;
  }
  int v;
  for (int i = 0; i < ve[u].size(); i++) {
    v = ve[u][i];
    if (v == par) continue;
    if (ve[v].size() >= 3) {
      st.push(v);
      dfs(v, u);
      if (!st.empty()) st.pop();
    }
  }
}
void dfs1(int u) {
  int v;
  for (int i = 0; i < ve[u].size(); i++) {
    v = ve[u][i];
    if (dist[v] != 0) continue;
    if (ve[v].size() == 1) {
      if (dist[u] != m) ok = 0;
    } else if (ve[v].size() == 2)
      ok = 0;
    else if (ve[v].size() == 3)
      ok = 0;
    else {
      dist[v] = dist[u] + 1;
      if (dist[v] > m)
        ok = 0;
      else
        dfs1(v);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  int u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    ve[u].push_back(v);
    ve[v].push_back(u);
  }
  u = -1;
  for (int i = 1; i <= n; i++) {
    if (ve[i].size() == 1) {
      u = ve[i][0];
      break;
    }
  }
  if (u == -1 || ve[u].size() < 3)
    ok = 0;
  else {
    st.push(u);
    dfs(u, 0);
    if (root == -1)
      ok = 0;
    else {
      dist[root] = 1;
      dfs1(root);
    }
  }
  if (!ok)
    cout << "NO" << '\n';
  else
    cout << "YES" << '\n';
}
