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;
const int maxn = 1e5 + 5;
vector<int> G[maxn];
int que[maxn], l, r, de[maxn], pre[maxn];
void add(int f, int t) { G[f].push_back(t); }
int bfs(int o) {
  l = r = 0;
  que[r++] = o;
  memset(de, -1, sizeof de);
  de[o] = 0;
  pre[o] = -1;
  while (l <= r) {
    int u = que[l++];
    for (int i = 0; i < G[u].size(); ++i) {
      int to = G[u][i];
      if (de[to] != -1) continue;
      pre[to] = u;
      de[to] = de[u] + 1;
      que[r++] = to;
    }
  }
  return que[r - 1];
}
bool check(int o, int fa, int k) {
  if (k == 0) {
    int son = 0;
    for (int i = 0; i < G[o].size(); ++i) {
      int to = G[o][i];
      if (to == fa) continue;
      son++;
    }
    if (son > 0) return false;
    return true;
  }
  int cnt = 0;
  for (int i = 0; i < G[o].size(); ++i) {
    int to = G[o][i];
    if (to == fa) continue;
    if (!check(to, o, k - 1)) return false;
    cnt++;
  }
  if (cnt < 3) return false;
  return true;
}
int main() {
  int n, k, u, v;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  u = bfs(1);
  v = bfs(u);
  if (de[v] % 2 != 0)
    puts("No");
  else {
    int center = v;
    for (int i = 0; i < de[v] / 2; ++i) center = pre[center];
    if (check(center, -1, k))
      puts("Yes");
    else
      puts("No");
  }
  return 0;
}
