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>
const long long LINF = 1e18;
const int INF = 1e9;
const int M = 1e9 + 7;
const double EPS = 1.0e-9;
const double PI = acos(-1.0);
using namespace std;
vector<int> g[100100];
int n;
int k;
int c[100100];
bool fl = 0;
int used[100100];
void reverseBfs() {
  queue<int> q;
  for (int i = 0; i < n; i++)
    if (g[i + 1].size() == 1) {
      q.push(i + 1);
      used[i + 1] = 1;
    }
  while (!q.empty()) {
    int v = q.front();
    int len = g[v].size();
    for (int i = 0; i < len; i++) {
      int cur = g[v][i];
      if (used[cur] < 2) {
        if (!used[cur]) q.push(cur);
        used[cur] = 1;
        if (c[cur] && c[cur] != c[v] + 1) fl = 1;
        c[cur] = c[v] + 1;
      }
    }
    q.pop();
    used[v] = 2;
    if (fl) break;
  }
}
int main() {
  std::ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  reverseBfs();
  int m = 0;
  for (int i = 0; i < n; i++) m = max(m, c[i + 1]);
  int cnt = 0;
  for (int i = 0; i < n; i++) cnt += c[i + 1] == m;
  fl |= cnt > 1;
  for (int i = 1; i < n + 1; i++) {
    if (c[i] && c[i] < m) fl |= g[i].size() < 4;
    if (c[i] == m) fl |= g[i].size() < 3;
  }
  fl || m != k ? cout << "No" : cout << "Yes";
  return 0;
}
