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;
inline void read(int &x) {
  char c = getchar();
  x = 0;
  while (c > '9' || c < '0') c = getchar();
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
}
const int N = 100005;
int d[N], n, K, x, y, in[N];
int WT[N];
vector<int> v[N], e;
int main() {
  read(n);
  read(K);
  if (K > n) {
    puts("No");
    return 0;
  }
  for (int i = 1; i < n; ++i) {
    read(x);
    read(y);
    ++d[x];
    ++d[y];
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 1; i <= n; ++i)
    if (d[x] == 1) {
      if (d[v[x][0]] == 1) {
        puts("No");
        return 0;
      }
    }
  for (int i = 1; i <= K; ++i) {
    for (int j = 1; j <= n; ++j)
      if (!in[j] && d[j] == 1) {
        in[j] = 1;
        e.push_back(j);
        if (v[j].size() < 4 && i != 1) {
          puts("No");
          return 0;
        }
      } else {
        if (!in[j] && WT[j]) {
          puts("No");
          return 0;
        }
      }
    if (e.size() == 0) {
      puts("No");
      return 0;
    }
    for (unsigned j = 0; j < e.size(); ++j)
      for (unsigned k = 0; k < v[e[j]].size(); ++k) {
        --d[v[e[j]][k]];
        WT[v[e[j]][k]] = 1;
      }
    e.clear();
  }
  int p = 0;
  for (int i = 1; i <= n; ++i)
    if (!in[i]) {
      if (p || d[i] != 0 || v[i].size() < 3) {
        puts("No");
        return 0;
      }
      p = 1;
    }
  puts(p ? "Yes" : "No");
  return 0;
}
