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, S, k, cnt[100010], flag, x, y;
int K, first[200010], last[200010], a[200010], ne[200010];
char ch;
void read(int &digit) {
  int QAQ;
  digit = 0;
  ch = getchar();
  while ((ch < '0' || ch > '9') && (ch != '-')) ch = getchar();
  if (ch == '-') {
    QAQ = -1;
    ch = getchar();
  } else
    QAQ = 1;
  while (ch >= '0' && ch <= '9') {
    digit = digit * 10 + ch - '0';
    ch = getchar();
  }
  digit = digit * QAQ;
}
void add(int x, int y) {
  K++;
  a[K] = y;
  if (first[x] == 0)
    first[x] = K;
  else
    ne[last[x]] = K;
  last[x] = K;
}
int dfs(int x, int fa, int d) {
  if (S != 0) return (0);
  int tmp = 0;
  int flag = 0;
  for (int i = first[x]; i; i = ne[i])
    if (a[i] != fa) {
      flag = 1;
      if (dfs(a[i], x, d + 1) == 1) tmp = 1;
    }
  if (tmp == 1 && d == k) S = x;
  if (flag == 0 && d == 2 * k) tmp = 1;
  return (tmp);
}
void dfs2(int x, int fa, int s) {
  if (flag == 0) return;
  if (s > k) {
    flag = 0;
    return;
  }
  int cnt = 0;
  int son = 1;
  for (int i = first[x]; i; i = ne[i])
    if (a[i] != fa) {
      son = 0;
      cnt++;
      dfs2(a[i], x, s + 1);
    }
  if (son == 1 && s != k) flag = 0;
  if (son == 0 && cnt < 3) flag = 0;
}
int main() {
  read(n);
  read(k);
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i < n; i++) {
    read(x);
    read(y);
    add(x, y);
    add(y, x);
    cnt[x]++;
    cnt[y]++;
  }
  S = 0;
  for (int i = 1; i <= n; i++)
    if (cnt[i] == 1) {
      dfs(i, 0, 0);
      break;
    }
  if (S == 0) {
    printf("No");
    return 0;
  }
  flag = 1;
  dfs2(S, 0, 0);
  if (flag == 0)
    printf("No");
  else
    printf("Yes");
  return 0;
}
