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 N = 2e5;
int n, k, cnt[N], was[N], was2[N];
vector<int> g[N];
vector<int> lvl[20];
void fail() {
  puts("No");
  exit(0);
}
int main() {
  cin >> n >> k;
  k++;
  if (k > 15) fail();
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i < n + 1; ++i) {
    if (g[i].empty()) fail();
    if (g[i].size() == 1) {
      lvl[k].push_back(i);
      was[i] = was2[i] = 1;
      int v = g[i][0];
      cnt[v]++;
      if (!was[v]) {
        was[v] = 1;
        lvl[k - 1].push_back(v);
      }
    }
  }
  for (int i = k - 1; i > 1; --i) {
    if (lvl[i].empty()) fail();
    for (auto v : lvl[i + 1]) was2[v] = 1;
    for (auto v : lvl[i]) {
      if (cnt[v] < 3) fail();
      int pre = -1;
      for (auto u : g[v]) {
        if (!was2[u]) {
          if (pre != -1) fail();
          pre = u;
        }
      }
      if (pre == -1) fail();
      cnt[pre]++;
      if (!was[pre]) {
        was[pre] = 1;
        lvl[i - 1].push_back(pre);
      }
    }
  }
  if (lvl[1].size() != 1) fail();
  if (g[lvl[1][0]].size() < 3) fail();
  puts("Yes");
}
