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 mod = 998244353;
const int MAXN = 1e5 + 5;
int add(int x, int y) {
  x += y;
  if (x >= mod) return x - mod;
  return x;
}
int mul(int x, int y) { return (long long)x * y % mod; }
int sub(int x, int y) {
  x -= y;
  if (x < 0) return x + mod;
  return x;
}
void fail() {
  cout << "No";
  exit(0);
}
vector<int> e[MAXN];
int dis[MAXN];
int ok[MAXN];
int k;
int dfs(int u, int v, int dub) {
  dis[v] = dub;
  int mx = dub;
  for (auto w : e[v]) {
    if (u != w) mx = max(mx, dfs(v, w, dub + 1));
  }
  if (mx == 2 * k) {
    ok[v] = 1;
  }
  return mx;
}
void check(int u, int v, int dub) {
  if (e[v].size() == 1) {
    if (dub != k) fail();
    return;
  }
  if (u != v) {
    if (e[v].size() < 4) fail();
  }
  for (auto w : e[v]) {
    if (u != w) check(v, w, dub + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs(0, 0, 0);
  int mx = 0, j = 0;
  for (int i = 0; i < n; ++i) {
    if (dis[i] > mx) {
      j = i;
      mx = dis[i];
    }
  }
  memset(ok, 0, sizeof ok);
  dfs(j, j, 0);
  mx = 0, j = 0;
  for (int i = 0; i < n; ++i) {
    if (dis[i] > mx) mx = dis[i];
  }
  if (mx != 2 * k) fail();
  j = -1;
  for (int i = 0; i < n; ++i) {
    if (dis[i] == mx / 2 && ok[i]) {
      if (j != -1) {
        fail();
      }
      j = i;
    }
  }
  if (e[j].size() < 3) fail();
  check(j, j, 0);
  cout << "Yes";
}
