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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void no() {
  cout << "No";
  exit(0);
}
const int N = 100100;
int n, k, deg[N], dist[N], center, par[N];
vector<int> g[N];
bool used[N];
void dfs(int v, int p = -1, int d = 0) {
  used[v] = true;
  dist[v] = d;
  par[v] = p;
  for (int to : g[v])
    if (!used[to]) dfs(to, v, d + 1);
}
void check(int v, int d = 0) {
  used[v] = true;
  if (deg[v] != 1 && v != center && deg[v] < 4) no();
  if (d > k) no();
  if (d == k && deg[v] != 1) no();
  if (deg[v] == 1 && d != k) no();
  for (int to : g[v])
    if (!used[to]) check(to, d + 1);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf(" %d%d", &a, &b);
    deg[a]++;
    deg[b]++;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  if (k > 10) no();
  int fin = -1, start = -1, mx = -1;
  for (int i = 1; i <= n; i++)
    if (deg[i] == 1) {
      start = i;
      break;
    }
  dfs(start);
  for (int i = 1; i <= n; i++)
    if (dist[i] > mx) {
      mx = dist[i];
      fin = i;
    }
  memset(used, 0, sizeof used);
  memset(dist, 0, sizeof dist);
  memset(par, 0, sizeof par);
  dfs(fin);
  center = start;
  for (int t = 0; t < k; t++) center = par[center];
  if (deg[center] < 3) no();
  memset(used, 0, sizeof used);
  check(center);
  cout << "Yes";
}
