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 + 10;
int n, k, degree[N], total[N];
set<int> G[N];
int main() {
  scanf("%d%d", &n, &k);
  if (n == 1) {
    return puts(!k ? "Yes" : "No"), 0;
  }
  for (register int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].insert(v);
    G[v].insert(u);
    ++degree[u], ++degree[v];
  }
  int nodes = n;
  for (register int i = 1;; ++i) {
    memset(total, 0, sizeof total);
    vector<int> s;
    for (register int j = 1; j <= n; ++j) {
      if (degree[j] == 1) {
        --nodes;
        int v = *G[j].begin();
        ++total[v];
        G[v].erase(j);
        G[j].erase(v);
        --degree[j];
        s.push_back(v);
      }
    }
    for (auto v : s) {
      --degree[v];
    }
    for (register int j = 1; j <= n; ++j) {
      if (total[j] > 0 && total[j] < 3) {
        return puts("No"), 0;
      }
    }
    if (nodes == 1) {
      return puts(i == k ? "Yes" : "No"), 0;
    }
  }
  return 0;
}
