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 INF = 0x3f3f3f3f;
const long long LLINF = 1e18;
const int maxN = 1e5 + 10;
int v, k;
int D[maxN];
vector<int> G[maxN];
queue<int> L;
int main() {
  scanf("%d %d", &v, &k);
  for (int i = 0; i < v - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--, b--;
    G[a].push_back(b);
    G[b].push_back(a);
    D[a]++, D[b]++;
  }
  for (int i = 0; i < v; i++) {
    if (D[i] == 1) L.push(i);
  }
  for (int i = 0; i < k; i++) {
    if (L.size() == 0) {
      printf("No\n");
      return 0;
    }
    map<int, int> mp;
    while (!L.empty()) {
      int x = L.front();
      L.pop();
      D[x] = 0;
      for (auto y : G[x]) {
        if (D[y] > 0) {
          D[y]--;
          mp[y]++;
        }
      }
    }
    for (auto p : mp) {
      if (p.second < 3) {
        printf("No\n");
        return 0;
      }
      L.push(p.first);
    }
  }
  if (L.size() == 1 && D[L.front()] == 0)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
