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 wh[100005];
int degree[100005];
int ty[100005];
vector<int> V[100005];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    V[a].push_back(b);
    V[b].push_back(a);
    ++degree[a];
    ++degree[b];
  }
  if (n <= 3) {
    cout << "No" << endl;
    return 0;
  }
  queue<int> Q;
  for (int i = 1; i <= n; ++i) {
    if (degree[i] == 1) {
      Q.push(i);
    }
  }
  int wo = -1;
  int coun = 0;
  while (!Q.empty()) {
    coun = 0;
    int u = Q.front();
    Q.pop();
    --degree[u];
    wo = u;
    bool flag = 1;
    int ok = -1;
    for (int j = 0; j < V[u].size(); ++j) {
      if (degree[V[u][j]] == 0 && ok == -1) {
        ++coun;
        ok = ty[V[u][j]];
      } else if (degree[V[u][j]] == 0) {
        ++coun;
        if (ok != ty[V[u][j]]) {
          flag = 0;
          break;
        }
      }
    }
    if (!flag || (coun && coun < 3)) {
      cout << "No" << endl;
      return 0;
    }
    ty[u] = ok + 1;
    for (int i = 0; i < V[u].size(); ++i) {
      if (degree[V[u][i]]) {
        --degree[V[u][i]];
        if (degree[V[u][i]] == 1) {
          Q.push(V[u][i]);
        }
      }
    }
  }
  if (ty[wo] == k)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}
