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;
namespace flyinthesky {
const int MAXN = 100000 + 5;
int zjlen, ct, st, ed, n, k, dis[MAXN], vis[MAXN];
vector<int> G[MAXN];
void dfs(int u, int fa) {
  dis[u] = dis[fa] + 1;
  for (int i = 0; i < (int)G[u].size(); ++i) {
    int v = G[u][i];
    if (v != fa) dfs(v, u);
  }
}
int dfs2(int u, int fa) {
  if (u == ed) {
    vis[ed] = 1;
    return true;
  }
  for (int i = 0; i < (int)G[u].size(); ++i) {
    int v = G[u][i];
    if (v != fa) {
      if (dis[v] == dis[u] + 1) {
        int gg = dfs2(v, u);
        if (gg) {
          vis[u] = 1;
          return true;
        }
      }
    }
  }
  return false;
}
int check(int u, int fa, int dep) {
  int dg = 0;
  for (int i = 0; i < (int)G[u].size(); ++i) {
    int v = G[u][i];
    if (v != fa) ++dg;
  }
  if (dep == zjlen / 2) {
    if (dg != 0) return false;
    return true;
  }
  if (dg < 3) return false;
  for (int i = 0; i < (int)G[u].size(); ++i) {
    int v = G[u][i];
    if (v != fa) {
      if (!check(v, u, dep + 1)) return false;
    }
  }
  return true;
}
void clean() { memset(vis, 0, sizeof vis), memset(dis, 0, sizeof dis); }
int solve() {
  cin >> n >> k;
  if (n == 1) return printf("No\n"), 0;
  clean();
  for (int u, v, i = 1; i < n; ++i)
    scanf("%d%d", &u, &v), G[u].push_back(v), G[v].push_back(u);
  dis[0] = 0, dfs(1, 0);
  st = 0;
  for (int i = 1; i <= n; ++i)
    if (dis[i] > dis[st]) st = i;
  memset(dis, 0, sizeof dis), dfs(st, 0);
  ed = 0;
  for (int i = 1; i <= n; ++i)
    if (dis[i] > dis[ed]) ed = i;
  zjlen = dis[ed] - 1;
  if (zjlen % 2 == 1) return printf("No\n"), 0;
  if (zjlen / 2 != k) return printf("No\n"), 0;
  ct = 0;
  dfs2(st, 0);
  for (int i = 1; i <= n; ++i)
    if (dis[i] - 1 == zjlen / 2 && vis[i]) {
      ct = i;
      break;
    }
  if (check(ct, 0, 0))
    return printf("Yes\n"), 0;
  else
    return printf("No\n"), 0;
  return 0;
}
}  // namespace flyinthesky
int main() {
  flyinthesky::solve();
  return 0;
}
