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>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
const int N = 100100;
vector<int> m[N];
int d[N], a[N];
void dfs(int i, int p) {
  a[i] = p;
  for (int j : m[i])
    if (j != p) {
      d[j] = d[i] + 1;
      dfs(j, i);
    }
}
bool dfs2(int i, int p, int k) {
  int c = 0;
  for (int j : m[i])
    if (j != p) c++;
  if (k == 0)
    return c == 0;
  else {
    if (c < 3) return 0;
    for (int j : m[i])
      if (j != p && !dfs2(j, i, k - 1)) return 0;
    return 1;
  }
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int i, j;
  for (int l = n - 1; l--;
       scanf("%d%d", &i, &j), i--, j--, m[i].push_back(j), m[j].push_back(i))
    ;
  i = 0;
  d[0] = 0;
  dfs(i, -1);
  for (int j = 0; j < n; j++)
    if (d[j] > d[i]) i = j;
  d[i] = 0;
  dfs(i, -1);
  for (int j = 0; j < n; j++)
    if (d[j] > d[i]) i = j;
  for (j = d[i] / 2; j > 0; i = a[i], j--)
    ;
  if (dfs2(i, -1, k))
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
