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 N = 1e5 + 4;
vector<int> adj[N];
int deg[N];
int lv[N];
int decr[N];
int vis[N];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    adj[x].push_back(y);
    adj[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  queue<pair<int, int>> q;
  bool yes = n > 3;
  for (int i = 1; i <= n; i++)
    if (deg[i] == 1) q.push({i, 0}), lv[i] = 1, decr[i] = 3;
  while (!q.empty()) {
    auto tmp = q.front();
    q.pop();
    int u = tmp.first;
    int p = tmp.second;
    vis[u] = 1;
    if (decr[u] < 3) yes = false;
    for (auto v : adj[u]) {
      if (vis[v]) continue;
      if (lv[v] != 0 && lv[v] != lv[u] + 1) {
        yes = false;
      }
      lv[v] = lv[u] + 1;
      deg[v]--;
      decr[v]++;
      if (lv[v] == k + 1 && deg[v] == 0)
        q.push({v, u});
      else if (deg[v] == 1)
        q.push({v, u});
    }
  }
  int mx = 0;
  for (int i = 1; i <= n; i++) mx = max(mx, lv[i]);
  yes &= (mx == k + 1);
  puts(yes ? "Yes" : "No");
}
