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;
int read() {
  int x;
  scanf("%d", &x);
  return x;
};
int n, k;
int dis[100050], d[100050];
int dp[100050];
vector<int> q[100050];
void dfs(int u, int fa) {
  for (int i = 0; i < q[u].size(); i++) {
    int v = q[u][i];
    if (v == fa) continue;
    dp[v] = dp[u] + 1;
    dfs(v, u);
  }
}
int mp;
bool dfs2(int u, int fa) {
  if (q[u].size() == 1 && dp[u] != mp) return false;
  if (dp[u] == mp) {
    if (q[u].size() == 1)
      return true;
    else
      return false;
  }
  int son = 0;
  for (int i = 0; i < q[u].size(); i++) {
    int v = q[u][i];
    if (v == fa) continue;
    son++;
    if (!dfs2(v, u)) return false;
  }
  if (son < 3) return false;
  return true;
}
int main() {
  n = read();
  k = read();
  memset(d, 0, sizeof(d));
  memset(dis, -1, sizeof(dis));
  for (int i = 1; i < n; i++) {
    int u = read();
    int v = read();
    d[v]++;
    d[u]++;
    q[u].push_back(v);
    q[v].push_back(u);
  }
  queue<int> qu;
  for (int i = 1; i <= n; i++) {
    if (d[i] == 1) {
      dis[i] = 0;
      qu.push(i);
    }
  }
  while (!qu.empty()) {
    int f = qu.front();
    qu.pop();
    for (int i = 0; i < q[f].size(); i++) {
      int v = q[f][i];
      if (dis[v] == -1 || dis[v] > dis[f] + 1) {
        dis[v] = dis[f] + 1;
        qu.push(v);
      }
    }
  }
  int maxx = 0, num = 0;
  int pos;
  for (int i = 1; i <= n; i++) {
    if (dis[i] == maxx)
      num++;
    else if (dis[i] > maxx) {
      maxx = dis[i];
      num = 1;
      pos = i;
    }
  }
  if (maxx == 0 || num > 1) {
    cout << "No" << endl;
  } else {
    dp[pos] = 1;
    dfs(pos, -1);
    mp = 0;
    for (int i = 1; i <= n; ++i) mp = max(mp, dp[i]);
    bool ans = dfs2(pos, -1);
    if (ans && mp == k + 1) {
      cout << "Yes" << endl;
    } else {
      cout << "No" << endl;
    }
  }
}
