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 INF = 1e9;
const int maxN = 100 * 1000 + 9;
int n, k;
vector<int> g[maxN];
void read() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
}
int layer[maxN];
void dfs(int u, int par, int h) {
  int childCnt = 0;
  if (g[u].size() == 1) {
    if (h != k) {
      cout << "No\n";
      exit(0);
    }
    return;
  }
  for (int i = 0, L = g[u].size(); i < L; i++) {
    int v = g[u][i];
    if (v == par) continue;
    childCnt++;
    dfs(v, u, h + 1);
  }
  if (childCnt < 3) {
    cout << "No\n";
    exit(0);
  }
}
void solv() {
  fill_n(layer, maxN, INF);
  queue<int> que;
  for (int i = 1; i <= n; i++) {
    if (g[i].size() == 1) {
      que.push(i);
      layer[i] = 0;
    } else if (g[i].size() < 3) {
      cout << "No\n";
      return;
    }
  }
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    for (int i = 0, L = g[u].size(); i < L; i++) {
      int v = g[u][i];
      if (layer[v] > layer[u] + 1) {
        layer[v] = layer[u] + 1;
        que.push(v);
      }
    }
  }
  int mx = *max_element(layer + 1, layer + n + 1);
  int cnt = count(layer + 1, layer + n + 1, mx);
  if (cnt > 1) {
    cout << "No\n";
    return;
  }
  if (cnt == 1) {
    if (mx != k) {
      cout << "No\n";
      return;
    }
    int root = 1;
    for (int i = 1; i <= n; i++)
      if (layer[i] == mx) root = i;
    dfs(root, 0, 0);
    cout << "Yes\n";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  read();
  solv();
  return 0;
}
