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;
int vis[100100];
int deg[100100];
int cnt[100100];
int dis[100100];
vector<int> lis[100100];
int main() {
  int i;
  scanf("%d%d", &n, &k);
  for (i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    lis[a].push_back(b);
    lis[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  memset(dis, -1, sizeof(dis));
  queue<int> q;
  for (i = 0; i < n; i++) {
    if (deg[i] == 1) {
      q.push(i);
      vis[i] = 1;
      dis[i] = 0;
    }
  }
  bool fl = 1;
  while (!q.empty()) {
    int here = q.front();
    if ((dis[here] && dis[here] != k && cnt[here] < 3) ||
        (dis[here] == k && cnt[here] < 2)) {
      fl = 0;
    }
    q.pop();
    for (int &there : lis[here]) {
      if (vis[there]) continue;
      if (~dis[there] && dis[there] != dis[here] + 1) {
        fl = 0;
      }
      dis[there] = dis[here] + 1;
      deg[there]--;
      if (deg[there] == 1) {
        q.push(there);
        vis[there] = 1;
      }
      cnt[there]++;
    }
  }
  int c = 0, d = 0;
  for (i = 0; i < n; i++) {
    if (dis[i] == k) c++;
    if (dis[i] > k) d++;
  }
  if (c != 1 || d > 0) fl = 0;
  puts(fl ? "Yes" : "No");
  return 0;
}
