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 GCC optimize("Ofast")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,mmx,avx,tune=native")
using namespace std;
const int MOD = 1000000007;
const int UNDEF = -1;
const int INF = 1 << 30;
template <typename T>
inline bool chkmax(T &aa, T bb) {
  return aa < bb ? aa = bb, true : false;
}
template <typename T>
inline bool chkmin(T &aa, T bb) {
  return aa > bb ? aa = bb, true : false;
}
vector<int> findTreeCenters(vector<vector<int> > tree) {
  int n = tree.size();
  vector<int> leaves;
  vector<int> degree(n);
  for (int i = 0; i < n; i++) {
    degree[i] = tree[i].size();
    if (degree[i] <= 1) {
      leaves.push_back(i);
    }
  }
  int removedLeaves = leaves.size();
  while (removedLeaves < n) {
    vector<int> nleaves;
    for (int u : leaves) {
      for (int v : tree[u]) {
        if (--degree[v] == 1) {
          nleaves.push_back(v);
        }
      }
    }
    leaves = nleaves;
    removedLeaves += leaves.size();
  }
  return leaves;
}
void fail() {
  printf("No\n");
  exit(0);
}
void succ() {
  printf("Yes\n");
  exit(0);
}
void must(bool cond) {
  if (!cond) fail();
}
vector<vector<int> > g;
void dfs(int x, int p, int k) {
  int numchild = g[x].size();
  if (p != -1) numchild--;
  if (k == 0) {
    must(numchild == 0);
    return;
  }
  if (numchild == 0) {
    must(k == 0);
    return;
  }
  must(numchild >= 3);
  for (auto &y : g[x]) {
    if (y != p) {
      dfs(y, x, k - 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  g.resize(n);
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  vector<int> centers = findTreeCenters(g);
  if (centers.size() != 1) fail();
  int center = centers[0];
  dfs(center, -1, k);
  succ();
}
