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;
vector<int> E[200000];
int d[200000];
int main() {
  int n, K;
  cin >> n >> K;
  for (int i = 0; i < (n - 1); i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    E[a].push_back(b);
    E[b].push_back(a);
  }
  queue<int> que;
  memset(d, -1, sizeof(d));
  for (int i = 0; i < (n); i++) {
    if (E[i].size() == 1) {
      d[i] = 0;
      que.push(i);
    }
  }
  while (!que.empty()) {
    int p = que.front();
    que.pop();
    for (int u : E[p]) {
      if (d[u] == -1) {
        d[u] = d[p] + 1;
        que.push(u);
      }
    }
  }
  int root = max_element(d, d + n) - d;
  memset(d, -1, sizeof(d));
  d[root] = 0;
  que.push(root);
  while (!que.empty()) {
    int p = que.front();
    que.pop();
    for (int u : E[p]) {
      if (d[u] == -1) {
        d[u] = d[p] + 1;
        que.push(u);
      }
    }
  }
  if (E[root].size() < 3) {
    puts("No");
    return 0;
  }
  for (int i = 0; i < (n); i++) {
    if (i == root) continue;
    if (E[i].size() == 1 && d[i] != K) {
      puts("No");
      return 0;
    }
    if (E[i].size() != 1 && E[i].size() <= 3) {
      puts("No");
      return 0;
    }
  }
  puts("Yes");
}
