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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int grand(int x) { return uniform_int_distribution<int>(0, x - 1)(rng); }
const int maxn = 1e6 + 7;
const long long inf = 1e15;
const long long md = 1e9 + 7;
const string NO = "No";
const string YES = "Yes";
long long gcd(long long x, long long y) {
  if (x < y) return gcd(y, x);
  if (y == 0) return x;
  return gcd(y, x % y);
}
vector<int> G[maxn];
vector<int> path;
int K;
bool ok = false;
int deg[maxn];
void dfs(int x, int pi, int m) {
  if (ok) return;
  path.push_back(x);
  if (path.size() == K + K + 1) {
    ok = true;
    return;
  }
  for (int y : G[x]) {
    if (y == pi) continue;
    dfs(y, x, m + 1);
  }
  path.pop_back();
}
void nope() {
  cout << NO << "\n";
  exit(0);
}
void yeah() {
  cout << YES << "\n";
  exit(0);
}
void go(int x, int pi, int h) {
  if (h == K) {
    if (deg[x] != 1) nope();
    return;
  }
  int ed = deg[x];
  if (pi > 0) ed--;
  if (ed < 3) nope();
  for (int y : G[x]) {
    if (y == pi) continue;
    go(y, x, h + 1);
  }
}
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
  };
  int n;
  cin >> n >> K;
  fill(deg, deg + maxn, 0);
  for (int j = 1; j < n; j++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  int z = -1;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      z = i;
      break;
    }
  }
  path.clear();
  dfs(z, -1, 1);
  if (!ok) {
    nope();
  }
  int v = path[K];
  go(v, -1, 0);
  yeah();
}
