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 mx = 1e5 + 5;
int n, k, kt;
vector<int> e[mx];
int center = -1;
bool sol = true;
int maxDist = -1;
void dfsFindDist(int x, int p, int dist) {
  maxDist = max(maxDist, dist);
  for (auto it : e[x]) {
    if (it == p) continue;
    dfsFindDist(it, x, dist + 1);
  }
}
bool mark[mx];
void dfsFind(int x) {
  stack<int> s;
  s.push(x);
  mark[x] = true;
  while (!s.empty()) {
    if (s.size() == maxDist + 1) {
      int t = maxDist / 2;
      while (t--) {
        s.pop();
      }
      center = s.top();
      return;
    }
    int x = s.top();
    for (auto it : e[x]) {
      if (!mark[it]) {
        s.push(it);
        mark[it] = true;
        break;
      }
    }
    if (s.top() == x) s.pop();
  }
}
void dfsCheck(int x, int p, int d) {
  if (d == k && e[x].size() != 1) {
    cout << "No\n";
    exit(0);
  };
  if (d != k && e[x].size() < 3 + (x != center)) {
    cout << "No\n";
    exit(0);
  };
  for (auto it : e[x]) {
    if (it == p) continue;
    dfsCheck(it, x, d + 1);
  }
}
int main() {
  cin >> n >> k;
  if (n < 4) {
    cout << "No\n";
    exit(0);
  };
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if (e[i].size() == 1) {
      dfsFindDist(i, 0, 0);
      if (maxDist % 2) {
        cout << "No\n";
        exit(0);
      };
      dfsFind(i);
      break;
    }
  }
  dfsCheck(center, 0, 0);
  cout << (sol ? "Yes" : "No") << endl;
  return 0;
}
