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 dist[100005], par[100005];
vector<int> vec[100005];
void dfs1(int s, int pp) {
  par[s] = pp;
  dist[s] = dist[pp] + 1;
  for (int v : vec[s]) {
    if (v == pp) continue;
    dfs1(v, s);
  }
}
bool dfs2(int s, int pp, int hn) {
  if (hn == 0) return (vec[s].size() == 1 && vec[s][0] == pp);
  int cnt = 0;
  int res = 1;
  for (int v : vec[s]) {
    if (v == pp) continue;
    cnt++;
    res &= dfs2(v, s, hn - 1);
  }
  return (res & (cnt >= 3));
}
void prexit() {
  printf("No\n");
  exit(0);
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    vec[x].push_back(y);
    vec[y].push_back(x);
  }
  int root = -1;
  for (int i = 1; i <= n; i++) {
    if (vec[i].size() == 1) {
      root = i;
      break;
    }
  }
  dist[0] = -1;
  dfs1(root, 0);
  int dm = 0, dmn = -1;
  for (int i = 1; i <= n; i++) {
    if (dist[i] > dm) dm = dist[i], dmn = i;
  }
  if (dm != 2 * k) prexit();
  for (int i = 1; i <= k; i++) dmn = par[dmn];
  int ans = dfs2(dmn, 0, k);
  if (ans)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
