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("O3")
using namespace std;
const long double EPS = 0.000000001;
const long long INF64 = 1000000000000000000;
const long long INF32 = 2000000000;
bool iss_pt_in_triangle(pair<long long, long long> a,
                        pair<long long, long long> b,
                        pair<long long, long long> c,
                        pair<long long, long long> point) {
  long long x1 = a.first, x2 = b.first, x3 = c.first, x0 = point.first;
  long long y1 = a.second, y2 = b.second, y3 = c.second, y0 = point.second;
  long long aa = (x1 - x0) * (y2 - y1) - (x2 - x1) * (y1 - y0);
  long long bb = (x2 - x0) * (y3 - y2) - (x3 - x2) * (y2 - y0);
  long long cc = (x3 - x0) * (y1 - y3) - (x1 - x3) * (y3 - y0);
  if (aa == 0 || bb == 0 || cc == 0) return 1;
  if (aa >= 0 && bb >= 0 && cc >= 0) return 1;
  if (aa <= 0 && bb <= 0 && cc <= 0) return 1;
  return 0;
}
bool iss_ll(long long &__1, long long &__2) { return INF64 / __1 >= __2; }
bool equal(long double _1, long double _2) { return (abs(_1 - _2) < EPS); }
bool iss_sqr(long long &__1) {
  long long __2 = sqrt((long double)__1);
  return __2 * __2 == __1;
};
bool iss_prime(long long __1) {
  for (long long i = 2; i * i <= __1; ++i)
    if (!(__1 % i)) return 0;
  return 1;
}
long long gcd(long long __1, long long __2) {
  while (__1 && __2)
    if (__1 > __2)
      __1 = __1 % __2;
    else
      __2 = __2 % __1;
  return __1 + __2;
}
long long lcm(long long &__1, long long &__2) {
  return __1 * __2 / (gcd(__1, __2));
}
long long sq_cube(long long x) {
  if (x == 0) return 1;
  long long l = 0, r = 2e6;
  while (l + 1 < r) {
    long long m = (l + r) / 2;
    if (m * m * m > x)
      r = m;
    else
      l = m;
  }
  return l;
}
set<long long> G[100100];
vector<long long> GR[100100];
bool used[100100];
void dfs(long long v) {
  if (used[v]) return;
  used[v] = 1;
  for (long long i : G[v]) dfs(i);
}
long long n, k;
long long step(long long k) {
  long long res = 1;
  for (long long i = 0; i < k; ++i) res *= 3;
  return res;
}
long long dfss(long long v, long long deep) {
  long long ans = deep;
  used[v] = 1;
  if (deep != k) {
    if ((GR[v].size() < 4 && deep > 0 && deep != k) ||
        (deep == 0 && GR[v].size() < 3)) {
      cout << "No"
           << "\n";
      exit(0);
    }
  }
  for (long long i : GR[v]) {
    if (used[i]) continue;
    ans = max(dfss(i, deep + 1), ans);
  }
  if (ans != k) {
    cout << "No"
         << "\n";
    exit(0);
  }
  return ans;
}
signed main(signed argc, const char *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  if (k > 20) {
    cout << "No"
         << "\n";
    return 0;
  }
  for (long long i = 1; i < n; ++i) {
    long long x, y;
    cin >> x >> y;
    G[x].insert(y);
    G[y].insert(x);
    GR[x].push_back(y);
    GR[y].push_back(x);
  }
  dfs(1);
  for (long long i = 1; i <= n; ++i) {
    if (!used[i]) {
      cout << "No"
           << "\n";
      return 0;
    }
  }
  vector<long long> VEC;
  long long k1 = k;
  long long pos = 0;
  for (long long i = 1; i <= n; ++i)
    if (G[i].size() == 1) VEC.push_back(i);
  while (k--) {
    if (VEC.size() == 0) {
      cout << "No"
           << "\n";
      return 0;
    }
    if (k == 0) {
      pos = *G[VEC.front()].begin();
    }
    for (long long i : VEC) {
      G[*G[i].begin()].erase(i);
      G[i].clear();
    }
    VEC.clear();
    for (long long i = 1; i <= n; ++i) {
      if (G[i].size() == 1) VEC.push_back(i);
      if (G[i].size() == 2) {
        cout << "No"
             << "\n";
        return 0;
      }
    }
  }
  memset(used, 0, sizeof(used));
  k = k1;
  dfss(pos, 0);
  cout << "Yes"
       << "\n";
}
