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 maxn = 1e5 + 10;
int d[maxn], t[maxn], n, k, maxk, root, mx[maxn], mn[maxn], sz[maxn];
bool flag[maxn];
vector<int> v[maxn];
queue<int> q;
void dfs(int dep, int u, int f = 0) {
  sz[u] = 1;
  if (v[u].size() > 1) mn[u] = 1e9;
  for (int i = 0; i < v[u].size(); i++) {
    if (v[u][i] == f) continue;
    dfs(dep + 1, v[u][i], u);
    sz[u] += sz[v[u][i]];
    mx[u] = max(mx[v[u][i]], mx[u]);
    mn[u] = min(mx[v[u][i]], mn[u]);
  }
  mx[u]++;
  mn[u]++;
}
bool dfs1(int dep, int u, int f = 0) {
  bool ret = 1;
  ret &= (mn[u] == mx[u]);
  if (dep == k)
    ret &= (sz[u] == 1);
  else if (u != root)
    ret &= (v[u].size() > 3);
  else
    ret &= (v[u].size() > 2);
  for (int i = 0; i < v[u].size(); i++)
    if (v[u][i] != f) ret &= dfs1(dep + 1, v[u][i], u);
  return ret;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1, tmp1, tmp2; i < n; i++) {
    scanf("%d%d", &tmp1, &tmp2);
    v[tmp1].push_back(tmp2);
    v[tmp2].push_back(tmp1);
    d[tmp1]++;
    d[tmp2]++;
  }
  for (int i = 1; i <= n; i++) {
    if (d[i] == 1) {
      flag[i] = 1;
      q.push(i);
      t[i] = 0;
    }
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < v[u].size(); i++) {
      if (t[v[u][i]] != 0 && (fabs(t[u] - t[v[u][i]]) > 1)) {
        puts("No");
        return 0;
      }
      if (flag[v[u][i]]) continue;
      if (t[v[u][i]] != 0 && t[v[u][i]] != t[u] + 1) {
        puts("No");
        return 0;
      }
      t[v[u][i]] = t[u] + 1;
      if (t[v[u][i]] > k) {
        puts("No");
        return 0;
      }
      if (t[v[u][i]] == k && root == 0) {
        root = v[u][i];
      }
      if (t[v[u][i]] == k && root != 0 && root != v[u][i]) {
        puts("No");
        return 0;
      }
      --d[v[u][i]];
      if (root != v[u][i] && d[v[u][i]] == 1) {
        q.push(v[u][i]);
        flag[v[u][i]] = 1;
      }
    }
  }
  if (root == 0) {
    puts("No");
    return 0;
  }
  dfs(0, root, 0);
  if (dfs1(0, root))
    puts("Yes");
  else
    puts("No");
}
