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;
struct Edge {
  int v, nxt;
} e[200010];
int tot;
int first[100010];
int d[100010];
void build(int u, int v) {
  e[++tot] = (Edge){v, first[u]};
  first[u] = tot;
  d[u]++;
  return;
}
int f[100010];
int s[100010];
queue<int> q;
bool bfs() {
  memset(s, 0x7f, sizeof(s));
  for (int i = 1; i <= n; i++)
    if (d[i] == 1) {
      q.push(i);
      s[i] = k;
    }
  bool flag = false;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (s[u] < 0) return false;
    if (s[u] == 0) {
      if (flag) return false;
      flag = true;
    }
    if (s[u] < k && f[u] < 3) return false;
    if (d[u] - f[u] > 1) return false;
    for (int i = first[u]; i; i = e[i].nxt) {
      f[e[i].v]++;
      if (s[e[i].v] > k) {
        s[e[i].v] = s[u] - 1;
        q.push(e[i].v);
      }
    }
  }
  return flag;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    build(u, v);
    build(v, u);
  }
  if (bfs())
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
