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;
const int MAXN = 1e5 + 5;
int N, K, D, X, Y;
int A[MAXN];
int B[MAXN];
vector<int> V[MAXN];
void dfs(int par, int now) {
  A[now] = A[par] + 1;
  for (int next : V[now]) {
    if (next == par) {
      continue;
    }
    dfs(now, next);
  }
  return;
}
void cek(int par, int now, int dep) {
  if (dep == 0) {
    if (V[now].size() != 1) {
      cout << "No\n";
      exit(0);
    }
    return;
  }
  if (V[now].size() < 4) {
    cout << "No\n";
    exit(0);
  }
  for (int next : V[now]) {
    if (next == par) {
      continue;
    }
    cek(now, next, dep - 1);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> K;
  A[0] = -1;
  for (int i = 1; i < N; i++) {
    cin >> X >> Y;
    V[X].push_back(Y);
    V[Y].push_back(X);
  }
  dfs(0, 1);
  for (int i = 1; i <= N; i++) {
    B[i] = A[i];
    if (A[i] > A[X]) {
      X = i;
    }
  }
  dfs(0, X);
  for (int i = 1; i <= N; i++) {
    if (A[i] > A[Y]) {
      Y = i;
    }
  }
  D = A[Y];
  if (D & 1) {
    cout << "No\n";
    return 0;
  }
  D >>= 1;
  for (int i = 1; i <= N; i++) {
    if (A[i] == K && B[X] - B[i] == K) {
      Y = i;
      V[Y].push_back(0);
    }
  }
  cek(0, Y, D);
  if (D != K) {
    cout << "No\n";
    return 0;
  }
  cout << "Yes\n";
  return 0;
}
