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;
vector<int> g[100001];
int s[100001], n, num, kk;
int en[100001];
void gs(int x, int p) {
  if (s[x] > s[num]) num = x;
  for (int y : g[x]) {
    if (y == p) continue;
    s[y] = s[x] + 1;
    gs(y, x);
  }
}
void gc(int x, int p) {
  if (s[x] == 2 * kk) en[x] = 1;
  for (int y : g[x]) {
    if (y == p) continue;
    s[y] = s[x] + 1;
    gc(y, x);
    en[x] |= en[y];
  }
  if (en[x] && s[x] == kk) num = x;
}
bool sol(int x, int k, int p) {
  if (k == 0) return g[x].size() == 1;
  int cnt = 0;
  for (int y : g[x]) {
    if (y == p) continue;
    if (!sol(y, k - 1, x)) return 0;
    cnt++;
  }
  if (cnt < 3) return 0;
  return 1;
}
int main() {
  int k, i, a, b;
  scanf("%d%d", &n, &k);
  kk = k;
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  gs(1, 0);
  s[num] = 0;
  gc(num, 0);
  if (sol(num, k, 0))
    printf("Yes");
  else
    printf("No");
  getchar();
  getchar();
}
