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;
struct edges {
  int to, next;
} edge[202020];
int first[101010], now;
void adde(int x, int y) {
  now++;
  edge[now].to = y;
  edge[now].next = first[x];
  first[x] = now;
}
int v[101010], cnt[101010], d[101010];
queue<int> q;
int main() {
  int n, k, x, y, ans = 0, p;
  scanf("%d%d", &n, &k);
  memset(d, 0, sizeof(d));
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    adde(x, y);
    adde(y, x);
    d[x]++;
    d[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (d[i] == 1) {
      q.push(i);
      v[i] = 0;
    }
  while (!q.empty()) {
    p = q.front();
    q.pop();
    ans = v[p];
    if (d[p] > 1 || (cnt[p] < 3 && v[p] != 0)) {
      printf("No");
      return 0;
    }
    for (int e = first[p]; e; e = edge[e].next)
      if (d[edge[e].to]) {
        d[edge[e].to]--;
        cnt[edge[e].to]++;
        if (v[edge[e].to] == 0) {
          v[edge[e].to] = v[p] + 1;
          q.push(edge[e].to);
        }
        if (v[edge[e].to] != v[p] + 1) {
          printf("No");
          return 0;
        }
      }
    d[p] = 0;
  }
  if (ans == k)
    printf("Yes");
  else
    printf("No");
  return 0;
}
