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;
bool parCheck(vector<vector<int> > &adjList, vector<int> &leaves, set<int> &par,
              vector<int> &deg) {
  for (int i = 0; i < leaves.size(); i++) {
    int u = leaves[i];
    par.insert(adjList[u][0]);
    if (deg[adjList[u][0]] < 4) return 0;
  }
  for (auto it = par.begin(); it != par.end(); it++) {
    int u = *it, si = 0;
    for (int i = 0; i < adjList[u].size(); i++) {
      int x = adjList[u][i];
      if (deg[x] != 1) si++;
    }
    if (si > 1) return 0;
  }
  return 1;
}
bool kMulti(vector<vector<int> > &adjList, vector<int> &deg,
            vector<int> &leaves, int k) {
  if (leaves.size() < 3) return 0;
  if (k == 1) {
    int p = adjList[leaves[0]][0];
    for (int i = 1; i < leaves.size(); i++) {
      int u = leaves[i];
      if (p != adjList[u][0]) return 0;
    }
    return 1;
  }
  set<int> par;
  if (!parCheck(adjList, leaves, par, deg)) return 0;
  leaves.clear();
  for (auto it = par.begin(); it != par.end(); it++) {
    int u = *it, v;
    for (int i = 0; i < adjList[u].size(); i++) {
      int x = adjList[u][i];
      if (deg[x] == 1)
        adjList[x].clear(), deg[x]--;
      else
        v = x;
    }
    adjList[u].clear();
    adjList[u].push_back(v);
    deg[u] = 1;
    leaves.push_back(u);
  }
  return kMulti(adjList, deg, leaves, k - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n, k;
  cin >> n >> k;
  vector<int> deg(n + 1, 0), leaves;
  vector<vector<int> > adjList(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adjList[u].push_back(v);
    adjList[v].push_back(u);
    deg[u]++, deg[v]++;
  }
  for (int i = 1; i <= n; i++)
    if (deg[i] == 1) leaves.push_back(i);
  if (kMulti(adjList, deg, leaves, k))
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}
