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;
int cnt[100100];
int d[100100];
queue<int> qu;
vector<int> a[100010];
int main() {
  int n, k, x, y, last, must_cen = 0;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    cnt[x]++;
    cnt[y]++;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if (cnt[i] == 1) {
      d[i] = 1;
      qu.push(i);
    } else if (cnt[i] == 2) {
      cout << "NO";
      exit(0);
    } else if (cnt[i] == 3) {
      if (must_cen) {
        cout << "No";
        exit(0);
      }
      must_cen = i;
    }
  }
  int dem = 0, ok;
  while (!qu.empty()) {
    dem++;
    x = qu.front();
    qu.pop();
    last = d[x];
    if (must_cen == x)
      ok = true;
    else
      ok = false;
    if (last != d[x]) {
    }
    for (int i = 0; i < a[x].size(); i++) {
      cnt[a[x][i]]--;
      if (d[a[x][i]] != 0 && d[a[x][i]] != d[x] + 1) {
        cout << "No";
        exit(0);
      } else
        d[a[x][i]] = d[x] + 1;
      if (cnt[a[x][i]] == 1) {
        qu.push(a[x][i]);
      }
    }
    d[x] += 2;
  }
  if (last - 1 == k && dem == n && (ok == 1 || must_cen == 0))
    cout << "Yes";
  else
    cout << "No";
}
