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 MAX_N = 1e5 + 200;
int n, limit, head[MAX_N], current, deg[MAX_N], cnt[MAX_N];
bool vis[MAX_N];
struct edge {
  int to, nxt;
} edges[MAX_N << 1];
void addpath(int src, int dst) {
  edges[current].to = dst, edges[current].nxt = head[src];
  head[src] = current++;
}
void bfs() {
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (!vis[i] && deg[i] == 1) q.push(i), vis[i] = true;
  if (q.empty()) puts("No"), exit(0);
  memset(cnt, 0, sizeof(cnt));
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = head[u]; i != -1; i = edges[i].nxt)
      --deg[edges[i].to], cnt[edges[i].to]++;
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i] && cnt[i] > 0 && cnt[i] < 3) puts("No"), exit(0);
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d%d", &n, &limit);
  for (int i = 1, u, v; i <= n - 1; i++)
    scanf("%d%d", &u, &v), addpath(u, v), addpath(v, u), deg[u]++, deg[v]++;
  for (int i = 1; i <= limit; i++) bfs();
  int valid = 0;
  for (int i = 1; i <= n; i++) valid += !vis[i];
  if (valid != 1) puts("No"), exit(0);
  puts("Yes");
  return 0;
}
