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;
using ll = long long;
using uint = unsigned int;
using rl = double;
const int MAXN = 100009;
struct Vertex {
  bool removed;
  int degremoved;
  vector<int> to;
  int dto;
} v[MAXN];
int n;
int Check() {
  if (n < 4) return -1;
  set<int> toCheck;
  queue<int> toRemove;
  int calck = 0;
  int sz = n;
  for (int i = 1; i <= n; i++) {
    if (v[i].to.size() == 1) {
      toRemove.push(i);
      v[i].dto = v[i].to.front();
    }
  }
  if (toRemove.empty()) return -1;
  while (sz != 1) {
    ;
    while (toRemove.empty() == false) {
      int top = toRemove.front();
      toRemove.pop();
      ;
      if (v[top].removed) return -1;
      v[top].removed = true;
      v[v[top].dto].degremoved++;
      toCheck.insert(v[top].dto);
      sz--;
    }
    for (int t : toCheck) {
      ;
      if (v[t].removed) return -1;
      if (sz != 1 && (int)v[t].to.size() != v[t].degremoved + 1) return -1;
      if (sz == 1 && (int)v[t].to.size() != v[t].degremoved) return -1;
      if (v[t].degremoved < 3) return -1;
      ;
      for (auto to : v[t].to) {
        if (v[to].removed == false) {
          v[t].dto = to;
          break;
        }
      }
      toRemove.push(t);
    }
    toCheck.clear();
    calck++;
  };
  return calck;
}
int main() {
  int k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    v[a].to.push_back(b);
    v[b].to.push_back(a);
  }
  int ck = Check();
  if (ck == k)
    printf("Yes\n");
  else
    printf("No\n");
}
