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 n, k;
vector<int> graph[100009];
int level[100009];
int pp[100009];
void dfs1(int u, int par) {
  for (int i = 0; i < graph[u].size(); i++) {
    int v = graph[u][i];
    if (v == par) continue;
    pp[v] = u;
    level[v] = level[u] + 1;
    dfs1(v, u);
  }
  return;
}
bool flag = 1;
void dfs2(int u, int par) {
  if (graph[u].size() == 1 && par != -1) {
    if (level[u] != k) flag = 0;
    return;
  }
  if (graph[u].size() < 4 && par != -1) flag = 0;
  if (par == -1 && graph[u].size() < 3) flag = 0;
  for (int i = 0; i < graph[u].size(); i++) {
    int v = graph[u][i];
    if (v == par) continue;
    level[v] = level[u] + 1;
    dfs2(v, u);
  }
}
int main() {
  scanf("%d %d", &n, &k);
  int x, y;
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &x, &y);
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  if (k >= n || n < 4) {
    cout << "No" << endl;
    return 0;
  }
  level[1] = 0;
  dfs1(1, -1);
  int mx = -1;
  for (int i = 1; i <= n; i++) {
    if (level[i] > mx) {
      mx = level[i];
      x = i;
    }
  }
  level[x] = 0;
  pp[x] = -1;
  dfs1(x, -1);
  mx = -1;
  for (int i = 1; i <= n; i++) {
    if (level[i] > mx) {
      mx = level[i];
      y = i;
    }
  }
  if (mx != 2 * k) {
    cout << "No" << endl;
    return 0;
  }
  while (level[y] != k) y = pp[y];
  level[y] = 0;
  dfs2(y, -1);
  if (flag)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}
