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 = 5e5 + 5, M = N << 1;
int n, k, u, v, edgenum, Head[N], Next[M], vet[M], deg[N], dis[N];
queue<int> que;
inline void add(int u, int v) {
  Next[++edgenum] = Head[u];
  Head[u] = edgenum;
  vet[edgenum] = v;
}
void dfs(int u, int fa, int d) {
  int cnt = 0;
  for (int e = Head[u]; e; e = Next[e]) {
    int v = vet[e];
    if (v ^ fa) cnt++, dfs(v, u, d + 1);
  }
  if (d < k && cnt < 3 || d == k && cnt || d > k) puts("No"), exit(0);
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++)
    scanf("%d%d", &u, &v), add(u, v), add(v, u), deg[u]++, deg[v]++;
  for (int i = 1; i <= n; i++)
    if (deg[i] == 1)
      que.push(i);
    else
      dis[i] = -1;
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    for (int e = Head[u]; e; e = Next[e]) {
      int v = vet[e];
      if (dis[v] == -1) dis[v] = dis[u] + 1, que.push(v);
    }
  }
  u = 0;
  for (int i = 1; i <= n; i++)
    if (dis[i] == -1 || dis[i] > k)
      return puts("No"), 0;
    else if (dis[i] == k)
      if (u)
        return puts("No"), 0;
      else
        u = i;
  dfs(u, 0, 0);
  return puts("Yes"), 0;
}
