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 = 1e5 + 5;
int n, k, dp[N], y, level[N], kol[N], mx;
vector<int> g[N], a;
void dfs(int x, int p) {
  if (p != 0) dp[x] = dp[p] + 1;
  if (dp[x] > dp[y]) y = x;
  for (int i = 0; i < g[x].size(); ++i) {
    if (g[x][i] != p) dfs(g[x][i], x);
  }
}
bool get_ans(int x, int p) {
  if (x == y) {
    a.push_back(x);
    return 1;
  }
  for (int i = 0; i < g[x].size(); ++i) {
    int y = g[x][i];
    if (y == p) continue;
    if (get_ans(y, x)) {
      a.push_back(x);
      return 1;
    }
  }
  return 0;
}
void go(int x, int p, int d) {
  mx = max(mx, d);
  level[x] = d;
  kol[x] = g[x].size();
  if (p != 0) kol[x]--;
  for (int i = 0; i < g[x].size(); ++i) {
    int y = g[x][i];
    if (y == p) continue;
    if (level[x] == level[y]) {
      cout << "No";
      exit(0);
    }
    go(y, x, d + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  if (n < 4) return cout << "No", 0;
  if (k > 10) return cout << "No", 0;
  y = 0;
  dfs(1, 0);
  int x = y;
  y = 0;
  dfs(x, 0);
  get_ans(x, 0);
  int m = a.size();
  if (m % 2 == 0) return cout << "No", 0;
  x = a[m / 2];
  go(x, 0, 1);
  if (mx != k + 1) return cout << "No", 0;
  for (int i = 1; i <= n; ++i) {
    if (level[i] != mx && kol[i] < 3) return cout << "No", 0;
  }
  cout << "Yes";
}
